A marriage counselor has traditionally seen that the proportion \( p \) of all married couples for whom her communication program can prevent divorce is 80%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 80% of married couples. In a random sample of 215 married couples who completed her program, 180 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.05 level of significance?Perform a one-tailed test. Then complete the parts below.Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) ## Finding the Value of the Test StatisticThe value of this test statistic is the z-value corresponding to the sample proportion under the assumption that \( H_0 \) is true. Here is the formula and calculation for determining the z-value:\[ z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{\frac{36}{43} - 0.80}{\sqrt{\frac{0.80(1 - 0.80)}{215}}} \approx 1.364 \]### Explanation:- **Sample Proportion (\(\hat{p}\))**: The sample proportion is represented as \(\frac{36}{43}\).- **Population Proportion (\(p\))**: The population proportion is given as 0.80.- **Sample Size (\(n\))**: The sample size is 215.The formula used to find the z-value is:\[ z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \]This equation subtracts the population proportion (\(p\)) from the sample proportion (\(\hat{p}\)) and divides by the standard error of the proportion, which is calculated as:\[ \sqrt{\frac{p(1-p)}{n}} \]By plugging in the given values:\[ \hat{p} = \frac{36}{43} \approx 0.8372 \]\[ p = 0.80 \]\[ n = 215 \]\[ z = \frac{0.8372 - 0.80}{\sqrt{\frac{0.80 \times (1 - 0.80)}{215}}} \]Simplifying the denominator:\[ \sqrt{\frac{0.80 \times 0.20}{215}} = \sqrt{\frac{0.16}{215}} = \sqrt{0.000744186} \approx 0.02728 \]Now, putting everything together:\[ z = \frac{0.8372 - 0.80}{0.02728} = \frac{0.0372}{0.02728} \approx 1.364 \]Thus, the z-value is approximately **1.364**. This z-value can be used to determine the statistical significance under

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 80%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 80% of married couples. In a random sample of 215 married couples who completed her program, 180 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.05 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. A. State the null hypothesis Hoand the alternative hypothesis H1. Ho: H1: B. Find the value of the test statistic. (Round to three or more decimal places.) C. Find the critical value. (Round to three or more decimal places.) D. Is there enough evidence to support the marriage counselor's claim that the proportion of married couples for whom her program…
A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 79%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 79% of married couples. In a random sample of 250 married couples who completed her program, 205 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor's claim at the 0.05 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H and the alternative hypothesis H₁. H₂ : D H₁ : 0 (b) Determine the type of test statistic to use. (Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) 0 (d) Find the p-value. (Round to three or more decimal places.) 0 (e) Is there enough…
A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 77%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 77% of married couples. In a random sample of 215 married couples who completed her program, 167 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.10 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H0 and the alternative hypothesis H1. H0: H1: (b) Determine the type of test statistic to use. ▼(Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the p-value. (Round to three…
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A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 78%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 78% of married couples. In a random sample of 240 married couples who completed her program, 189 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.10 level of significance?Perform a one-tailed test. Then complete the parts below.
In a previous poll, 38% of adults with children under the age of 18 reported that their family ate dinner together seven nights. suppose that, in a more recent poll, 399 of 1099 adults with children under the age of 18 reported that their family ate dinner together seven nights a week. is there sufficient evidence that the proportion of families with children under the age of 18 who eat diner together seven nights a week has decreased? use the a=0.05 significance level
The Substance Abuse and Mental Health Services Administration (SAMHSA, 2017) estimates that 10.9% of the population of the United States age 18–24 had an episode of depression in the previous 12 months. Suppose a researcher takes a random sample of 225 United States adults aged 18–24. Determine the value of the sample proportion, p, such that 5% of samples of 225 United States adults age 18–24 are greater than p. You may find software or a z-table useful. Give your answer precise to three decimal places. p = II
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A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 75%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 75% of married couples. In a random sample of 245 married couples who completed her program, 185 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor's claim at the 0.01 level of significance? (a) State the null hypothesis Ho and the alternative hypothesis H₁. Ho Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) H₁ (b) Determine the type of test statistic to use. (Choose one) ▼ (c) Find the value of the test statistic. (Round to three or more decimal places.) 0 (d) Find the p-value. (Round to three or more decimal places.) (e) Is there enough…
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