Hypothesis Tests

$ (a.) \\[3ex] Two-tailed\:\:Hypothesis\:\:Test \\[3ex] p-value = 0.0888 \\[3ex] \alpha = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] 0.0888 \gt 0.05 \\[3ex] p-value \gt \alpha \\[3ex] $ Decision: Fail to reject the null hypothesis

$ (b.) \\[3ex] Two-tailed\:\:Hypothesis\:\:Test \\[3ex] p-value = 0.0888 \\[3ex] \alpha = 10\% = \dfrac{10}{100} = 0.1 \\[5ex] 0.0888 \le 0.1 \\[3ex] p-value \le \alpha \\[3ex] $ Decision: Reject the null hypothesis

$ (c.) \\[3ex] One-tail\:(Right-tailed)\:\:Hypothesis\:\:Test \\[3ex] p-value = 0.0888 \\[3ex] \alpha = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] 0.0888 \gt 0.05 \\[3ex] p-value \gt \alpha \\[3ex] $ Decision: Fail to reject the null hypothesis

$ H_0:\;\; p = 97\% = 0.97 \\[3ex] q = 1 - p = 1 - 0.97 = 0.03 \\[5ex] n = 1217 \\[3ex] \hat{p} = 89\% = 0.89 \\[3ex] z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{pq}{n}}} \\[9ex] z = \dfrac{0.89 - 0.97}{\sqrt{\dfrac{0.97(0.03)}{1217}}} \\[9ex] z = -16.36020652 \\[3ex] z \approx -16.36 $

p = 30% = 0.3
(a.) The null and alternative hypotheses are:

$ H_0:\;\; p = 0.3 \\[3ex] H_1:\;\; p \ne 0.3 \\[3ex] $ (b.) The test statistic, z = 0.6172
(c.) The probability value, p = 0.5371

$ H_0:\;\; \mu = 124\;mmHg \\[3ex] n = 279 \\[3ex] \bar{x} = 123.93\;mmHg \\[3ex] s = 15.71\;mmHg \\[3ex] t = \dfrac{\bar{x} - \mu}{\dfrac{s}{\sqrt{n}}} \\[9ex] t = \dfrac{123.93 - 124}{\dfrac{15.71}{\sqrt{279}}} \\[9ex] t = -0.0744258763 \\[3ex] t \approx -0.07 $

p = 32% = 0.32
(a.) The null and alternative hypotheses are:

$ H_0:\;\; p = 0.32 \\[3ex] H_1:\;\; p \lt 0.32 \\[3ex] $ (b.) The test statistic, z = -1.9174
(c.) The probability value, p = 0.0276

$ H_0:\;\; \sigma = 10bpm \\[3ex] n = 174 \\[3ex] s = 8.6bpm \\[3ex] \chi^2 = \dfrac{s^2(n - 1)}{\sigma^2} \\[5ex] \chi^2 = \dfrac{8.6^2(174 - 1)}{10^2} \\[5ex] \chi^2 = 127.9508 \\[3ex] \chi^2 \approx 127.95 $

$ (a.) \\[3ex] x = 115 \\[3ex] n = 200 \\[3ex] \hat{p} = \dfrac{x}{n} = \dfrac{115}{200} = 0.575 \\[5ex] (b.) \\[3ex] p_0 = 0.57 \\[3ex] (c.) \\[3ex] p = p_0 = 0.57 \\[3ex] q = 1 - p = 1 - 0.57 = 0.43 \\[3ex] z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{pq}{n}}} \\[9ex] z = \dfrac{0.575 - 0.57}{\sqrt{\dfrac{0.57 * 0.43}{200}}} \\[9ex] z = 0.1428279973 \\[3ex] z \approx 0.14 \\[3ex] $ The value of the test statistic tells that the observed proportion of students who passed introductory chemistry was 0.14 standard errors above the null hypothesis proportion of students.

0.1% = 10%
The​ p-value is the probability that if the null hypothesis is​ true (the person is guessing), a test statistic (p) will have a value as extreme as or more extreme than the observed value (gets 14 or more right).
So, the correct option is: D.: The probability that a person will get 14 or more​ right, if the person is truly​ guessing, is about 10​%.

We are only interested in the number of adults who regularly use supplemental vitamins

$ H_0:\:\: p = 0.47 \\[3ex] H_1:\:\: p \ne 0.47 \\[3ex] x = 91 \\[3ex] n = 170 \\[3ex] \hat{p} = \dfrac{x}{n} \\[5ex] \hat{p} = \dfrac{91}{170} \\[5ex] \hat{p} = 0.5352941176 \\[3ex] p = 0.47 \\[3ex] q = 1 - p \\[3ex] q = 1 - 0.47 \\[3ex] q = 0.53 \\[3ex] z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{pq}{n}}} \\[9ex] z = \dfrac{0.5352941176 - 0.47}{\sqrt{\dfrac{0.47 * 0.53}{170}}} \\[9ex] z = \dfrac{0.06529411765}{\sqrt{0.001465294118}} \\[5ex] z = \dfrac{0.06529411765}{0.03827916036} \\[5ex] z = 1.705735367 \\[3ex] \alpha = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] \dfrac{\alpha}{2} = \dfrac{0.05}{2} = 0.025 \\[5ex] \underline{Classical\:\:Approach} \\[3ex] -z_{\dfrac{\alpha}{2}} = -1.9599639861189817 \\[5ex] z_{\dfrac{\alpha}{2}} = 1.9599639861189817 \\[5ex] -z_{\dfrac{\alpha}{2}} \lt z \lt z_{\dfrac{\alpha}{2}} \\[5ex] -1.96 \lt 1.71 \lt 1.96 \\[3ex] $ It does not fall in the critical region
Decision: Do not reject the null hypothesis

$ \underline{P-value\:\:Approach} \\[3ex] P(z) = P(z \lt -1.705735367) + P(z \gt 1.705735367) \\[3ex] P(z) = 0.04402868083136646 + 0.04402868083136646 \\[3ex] P(z) = 0.08805736166 \\[3ex] P(z) \gt \alpha \\[3ex] 0.0881 \gt 0.05 \\[3ex] $ Decision: Do not reject the null hypothesis

Conclusion: There is insufficient evidence to warrant the rejection of the claim that forty seven percent of the adult population regularly uses supplemental vitamins.




(f.)
To change the conclusion, there should be sufficient evidence to warrant the rejection of the claim that forty seven percent of the adult population regularly uses supplemental vitamins.
This means that there should sufficient evidence to warrant the rejection of the null hypothesis.
This means that we have to change the decision as well (the decision led to the conclusion)
This means that we have to reject the null hypothesis
The easiest way to do this is to use the P-value Approach
For Two-tailed tests, we reject the null hypothesis if $P-value \le \alpha$
This means that we have to find a significance level that is greater than or equal to $P(z)$
$P(z) = 0.08805736166$
The common significance level that is greater than $0.08805736166$ is $0.1$
$0.1 = 10\%$
Therefore, the significance level that will change the conclusion is $10\%$

Let us verify with the calculator.




Student: Are you saying that if we test the hypothesis using a $1\%$ level of significance, the conclusion will be the same?
Teacher: That is correct.
Of the three most common significant levels, only the $10\%$ level of significance will give a different conclusion.

$ H_0:\:\: p = 0.45 \\[3ex] H_1:\:\: p \gt 0.45 \\[3ex] x = 194 \\[3ex] n = 360 \\[3ex] \hat{p} = \dfrac{x}{n} \\[5ex] \hat{p} = \dfrac{194}{360} \\[5ex] \hat{p} = 0.5388888889 \\[3ex] p = 0.45 \\[3ex] q = 1 - p \\[3ex] q = 1 - 0.45 \\[3ex] q = 0.55 \\[3ex] z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{pq}{n}}} \\[9ex] z = \dfrac{0.5388888889 - 0.45}{\sqrt{\dfrac{0.45 * 0.55}{360}}} \\[9ex] z = \dfrac{0.08888888889}{\sqrt{0.0006875}} \\[5ex] z = \dfrac{0.08888888889}{0.0262202212} \\[5ex] z = 3.390089206 \\[3ex] \alpha = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] \dfrac{\alpha}{2} = \dfrac{0.05}{2} = 0.025 \\[5ex] \underline{Classical\:\:Approach} \\[3ex] z_{\dfrac{\alpha}{2}} = 1.9599639861189817 \\[5ex] z \gt z_{\dfrac{\alpha}{2}} \\[5ex] 3.39 \gt 1.96 \\[3ex] $ It falls in the critical region
Decision: Reject the null hypothesis

$ \underline{P-value\:\:Approach} \\[3ex] P(z \gt 3.390089206) = 0.000349349418287237 \\[3ex] P(z \gt 3.390089206) \le \alpha \\[3ex] 0.0003 \le 0.05 \\[3ex] $ Decision: Reject the null hypothesis

Conclusion: There is sufficient evidence to support the claim that the initiative is working to increase retention.

$ (a.) \\[3ex] x = 39 \\[3ex] n = 100 \\[3ex] \hat{p} = \dfrac{x}{n} = \dfrac{39}{100} = 0.39 \\[5ex] (b.) \\[3ex] p_0 = 0.34 \\[3ex] (c.) \\[3ex] p = p_0 = 0.34 \\[3ex] q = 1 - p = 1 - 0.34 = 0.66 \\[3ex] z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{pq}{n}}} \\[9ex] z = \dfrac{0.39 - 0.34}{\sqrt{\dfrac{0.34 * 0.66}{100}}} \\[9ex] z = 1.055500827 \\[3ex] z \approx 1.06 \\[3ex] $ The value of the test statistic tells that the observed proportion of prisoners was 1.06 standard errors above the null hypothesis proportion of prisoners.

(a.) 72% = 0.72
Null hypotheses: The passing rate for introductory chemistry is 72​%.
Alternative hypotheses: A college chemistry instructor thinks the use of embedded tutors will improve the success rate in introductory chemistry courses.

$ H_0:\;\; p = 0.72 \\[3ex] H_1:\;\; p \gt 0.72 \\[3ex] $ (b.) The​ p-value is the probability (0.0091) that if the null hypothesis is​ true (the passing rate for introductory chemistry is 72​%), a test statistic (z = 2.36) will have a value as extreme as or more extreme than the observed value (will greatly improve the success rate in introductory chemistry courses).
In the context of the question, this implies that:
The probability of 159 or more introductory chemistry students passing out of a random sample of 200 students is 0.0091​, assuming the population proportion is 0.72.

(c.) In this​ situation, the​ p-value is ​small (less than 0.05), so the instructor should believe the success rate has improved.

$\hat{p_1}$ is the first sample proportion (proportion of urban families who regularly eat lunch together)
$\hat{q_1}$ is the complement of the first sample proportion
$\hat{p_2}$ is the second sample proportion (proportion of rural families who regularly eat lunch together)
$\hat{q_2}$ is the complement of the second sample proportion
$\overline{p}$ is the pooled sample proportion
$\overline{q}$ is the complement of the pooled sample proportion

$ H_0:\:\: p_1 = p_2 \:\:OR\:\: p_1 - p_2 = 0 \\[3ex] H_1:\:\: p_1 \lt p_2 \\[3ex] \hat{p_1} = \dfrac{x_1}{n_1} \\[5ex] \hat{p_1} = \dfrac{12}{75} \\[5ex] \hat{p_1} = 0.16 \\[3ex] \hat{p_2} = \dfrac{x_2}{n_2} \\[5ex] \hat{p_2} = \dfrac{40}{140} \\[5ex] \hat{p_2} = 0.2857142857 \\[3ex] 0.16 \lt 0.2857142857 \\[3ex] $ Yes, these sample proportions show what the researcher expected.
The sample proportion of urban families who regularly eat lunch together is lower than the sample proportion of rural families who regularly eat lunch together because $0.16 \lt 0.2857142857$

$ \overline{p} = \dfrac{x_1 + x_2}{n_1 + n_2} \\[5ex] \overline{p} = \dfrac{12 + 40}{75 + 140} \\[5ex] \overline{p} = \dfrac{52}{215} \\[5ex] \overline{p} = 0.2418604651 \\[3ex] \overline{q} = 1 - 0.2418604651 \\[3ex] \overline{q} = 0.7581395349 \\[3ex] z = \dfrac{(\hat{p_1} - \hat{p_2}) - (p_1 - p_2)}{\sqrt{\dfrac{\overline{p} * \overline{q}}{n_1} + \dfrac{\overline{p} * \overline{q}}{n_2}}} \\[10ex] z = \dfrac{(0.16 - 0.2857142857) - 0}{\sqrt{\dfrac{0.2418604651 * 0.7581395349}{75} + \dfrac{0.2418604651 * 0.7581395349}{140}}} \\[10ex] z = \dfrac{-0.1257142857}{\sqrt{\dfrac{0.1833639805}{75} + \dfrac{0.1833639805}{140}}} \\[10ex] z = \dfrac{-0.1257142857}{\sqrt{0.002444853074 + 0.001309742718}} \\[7ex] z = \dfrac{-0.1257142857}{\sqrt{0.003754595792}} \\[7ex] z = \dfrac{-0.1257142857}{0.06127475656} \\[7ex] z = -2.05164888 \\[3ex] \alpha = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] \dfrac{\alpha}{2} = \dfrac{0.05}{2} = 0.025 \\[5ex] \underline{Classical\:\:Approach} \\[3ex] -z_{\dfrac{\alpha}{2}} = -1.9599639861189817 \\[5ex] z \lt -z_{\dfrac{\alpha}{2}} \\[5ex] -2.05164888 \lt -1.9599639861189817 \\[3ex] $ It falls in the critical region
Decision: Reject the null hypothesis

$ \underline{P-value\:\:Approach} \\[3ex] P(z) = P(z \lt -2.05164888) P(z \lt -2.05164888) = 0.020101899077793972 \\[3ex] P(z \lt -2.05164888) \le \alpha \\[3ex] 0.020101899077793972 \le 0.05 \\[3ex] $ Decision: Reject the null hypothesis

Conclusion: There is sufficient evidence to support the claim that the sample proportion of urban families who regularly eat lunch together is lower than the sample proportion of rural families who regularly eat lunch together.




(g.)

$ Test\:\:statistic:\:\: z = -2.05164888 \approx -2.05 \\[3ex] P-value:\:\: P(z) = 0.020101899077793972 \approx 2\% \\[3ex] $

(a.) If the population proportion, p is not given, then assume p = 50% = 0.5
50% of 30 trials = 0.5(30) = 15
The person should get 15 trials right.

$ (b.) \\[3ex] \underline{Person\;A} \\[3ex] x = 21 \\[3ex] n = 30 \\[3ex] \hat{p} = \dfrac{x}{n} = \dfrac{21}{30} = 0.7 \\[5ex] \underline{Person\;B} \\[3ex] x = 19 \\[3ex] n = 30 \\[3ex] \hat{p} = \dfrac{x}{n} = \dfrac{19}{30} = 0.6333333333 \\[5ex] $ 0.7 > 0.6333333333; therefore
Person A will have a smaller​ p-value because that person's number of successes is further from the hypothesized number of successes

(a.) Number of African Americans the researcher would expect in a jury pool of 200 people is:

$ 24\% \;\;of\;\; 200 \\[3ex] = 0.24(200) \\[3ex] = 48\;\;African\;\;Americans \\[3ex] $ (b.) Expected hypothesized number of successes: 48
Pool A number of succcesses: 35 African American
Pool B number of successes: 26 African American

Which of them is further from 48?

|48 - 35| = |13| = 13
|48 - 26| = |22| = 22
13 > 12
Pool B will have a smaller​ p-value because that pool's number of African American people is further from the hypothesized number.

$ (a.) \\[3ex] x = 14 \\[3ex] n = 80 \\[3ex] \hat{p} = \dfrac{x}{n} = \dfrac{14}{80} = 0.175 \\[5ex] $ (b.) 18% = 0.18
Null hypotheses: Nationally the readmission rate for patients with pneumonia is 18​%.
Alternative hypotheses: A hospital was interested in knowing whether their readmission rate for pneumonia was less than the national percentage.

$ H_0:\;\; p = 0.18 \\[3ex] H_1:\;\; p \lt 0.18 \\[3ex] (c.) \\[3ex] q = 1 - p = 1 - 0.18 = 0.82 \\[3ex] z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{pq}{n}}} \\[9ex] z = \dfrac{0.175 - 0.18}{\sqrt{\dfrac{0.18 * 0.82}{80}}} \\[9ex] z = -0.1164050493 \\[3ex] z \approx -0.12 \\[3ex] $ This implies that:
The value of the test statistic tells that the observed proportion of readmissions was 0.12 standard errors below the null hypothesis proportion of readmissions. (it is below because of the negative sign)

(d.) The​ p-value is the probability (0.45) that if the null hypothesis is​ true (Nationally the readmission rate for patients with pneumonia is 18​%.), a test statistic (z = -0.12) will have a value as extreme as or more extreme than the observed value (their readmission rate for pneumonia was very less than the national percentage).
In the context of the question, this implies that:
The probability of getting 14 or fewer readmissions for pneumonia of a random sample of 80 patients with pneumonia is 0.45, assuming the population proportion is 0.18.
In this​ situation, the​ p-value is not small (because it is greater than 0.05), which indicates that the null hypothesis should not be doubted.

(a.) The null hypothesis is p = 0.20, where p is the probability of a correct answer.
The null hypotheses is true (or assumed to be true)
It makes sense that the probability of getting a correct answer by guessing is 20% so we assume the null hypothesis to be true
The test statistic compares the observed outcome with the outcome of the null hypothesis (expected outcome).
For this case, the observed outcome is close to the expected outcome.
Hence, the z-test statistic will be close to 0.

(b.) Because the null hypothesis is true, obtaining an unusual result is not likely.
Hence, the p-value will not be small.