# STA 3163-Assume that the data has a normal distribution

Question 1Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis. = 0.05 for a two-tailed test.±2.5751.764±1.96±1.6455 points Question 2Find the value of the test statistic z using z = A claim is made that the proportion of children who play sports is less than 0.5, and the sample statistics include n = 1671 subjects with 30% saying that they play a sport.3.3816.35-33.38-16.355 points Question 3Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). The test statistic in a right-tailed test is z = 0.52.0.6030; fail to reject the null hypothesis0.3015; fail to reject the null hypothesis0.3015; reject the null hypothesis0.0195; reject the null hypothesis5 points Question 4Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). The test statistic in a two-tailed test is z = -1.63.0.1032; fail to reject the null hypothesis0.0516; reject the null hypothesis0.0516; fail to reject the null hypothesis0.9484; fail to reject the null hypothesis5 points Question 5Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim. A skeptical paranormal researcher claims that the proportion of Americans that have seen a UFO, p, is less than 2 in every ten thousand. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is failure to reject the null hypothesis, state the conclusion in nontechnical terms.There is sufficient evidence to support the claim that the true proportion is less than 2 in ten thousand.There is sufficient evidence to support the claim that the true proportion is greater than 2 in ten thousand.There is not sufficient evidence to support the claim that the true proportion is greater than 2 in ten thousand.There is not sufficient evidence to support the claim that the true proportion is less than 2 in ten thousand.5 points Question 6Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test. A medical researcher claims that 6% of children suffer from a certain disorder. Identify the type I error for the test.Reject the claim that the percentage of children who suffer from the disorder is different from 6% when that percentage really is different from 6%.Reject the claim that the percentage of children who suffer from the disorder is equal to 6% when that percentage is actually 6%.Fail to reject the claim that the percentage of children who suffer from the disorder is equal to 6% when that percentage is actually 6%.Fail to reject the claim that the percentage of children who suffer from the disorder is equal to 6% when that percentage is actually different from 6%.5 points Question 7Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test. A cereal company claims that the mean weight of the cereal in its packets is 14 oz. Identify the type I error for the test.Reject the claim that the mean weight is 14 oz when it is actually greater than 14 oz.Fail to reject the claim that the mean weight is 14 oz when it is actually different from 14 oz.Reject the claim that the mean weight is 14 oz when it is actually 14 oz.Reject the claim that the mean weight is different from 14 oz when it is actually 14 oz.5 points Question 8Find the P-value for the indicated hypothesis test. In a sample of 47 adults selected randomly from one town, it is found that 9 of them have been exposed to a particular strain of the flu. Find the P-value for a test of the claim that the proportion of all adults in the town that have been exposed to this strain of the flu is 8%.0.00480.00240.02620.05245 points Question 9Find the critical value or values of based on the given information. H0: = 8.0 n = 10 = 0.012.088, 21.6661.735, 23.58923.20921.6665 points Question 10Find the critical value or values of based on the given information. H1: < 0.14 n = 23 = 0.1014.04214.848-30.81330.8135 points Question 11Find the number of successes x suggested by the given statement. Among 660 adults selected randomly from among the residents of one town, 30.2% said that they favor stronger gun-control laws.2001971991985 points Question 12Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2, Use the given sample sizes and numbers of successes to find the pooled estimate Round your answer to the nearest thousandth. n1 = 100;n2 = 100 x1 = 32;x2 = 330.2930.2270.3580.3255 points Question 13Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2. Use the given sample sizes and numbers of successes to find the z test statistic for the hypothesis test. n1 = 155; n2 = 146 x1 = 68; x2 = 59z = 7.466z = 0.435z = 0.607z = 13.8655 points Question 14Solve the problem. The table shows the number of pitchers with E.R.A’s below 3.5 in a random sample of sixty pitchers from the National League and in a random sample of fifty-two pitchers from the American League. Assume that you plan to use a significance level of alpha = 0.05 to test the claim that Find the critical value(s) for this hypothesis test. Do the data support the claim that the proportion of National League pitchers with an E.R.A. below 3.5 differs from the proportion of American League pitchers with an E.R.A. below 3.5?z = ±1.645; yesz = 1.645; yesz = ±2.575; noz = ±1.96; no5 points Question 15Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2. Use the given sample sizes and numbers of successes to find the P-value for the hypothesis test. n1 = 100;n2 = 100 x1 = 38;x2 = 400.04120.16100.77180.21305 points Question 16Construct the indicated confidence interval for the difference between population proportions p1 – p2. Assume that the samples are independent and that they have been randomly selected. x1 = 22, n1 = 38 and x2 = 31, n2 = 52; Construct a 90% confidence interval for the difference between population proportions p1 – p2.0.406 < p1 – p2 < 0.752-0.190 < p1 – p2 < 0.1560.373 < p1 – p2 < 0.7850.785 < p1 – p2 < 0.3735 points Question 17Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Two types of flares are tested and their burning times are recorded. The summary statistics are given below. Construct a 95% confidence interval for the differences between the mean burning time of the brand X flare and the mean burning time of the brand Y flare.3.5 min < muX – muY < 5.1 min3.2 min < muX – muY < 5.4 min3.6 min < muX – muY < 5.0 min3.8 min < muX – muY < 4.8 min5 points Question 18State what the given confidence interval suggests about the two population means. A paint manufacturer made a modification to a paint to speed up its drying time. Independent simple random samples of 11 cans of type A (the original paint) and 9 cans of type B (the modified paint) were selected and applied to similar surfaces. The drying times, in hours, were recorded. The summary statistics are as follows. The following 98% confidence interval was obtained for mu1 – mu2, the difference between the mean drying time for paint cans of type A and the mean drying time for paint cans of type B: 4.70hrs < mu1 – mu2 < 17.30hrs What does the confidence interval suggest about the population means?The confidence interval includes only positive values which suggests that the two population means might be equal. There doesn’t appear to be a significant difference between the mean drying time for paint type A and the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.The confidence interval includes only positive values which suggests that the mean drying time for paint type A is smaller than the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.The confidence interval includes 0 which suggests that the two population means might be equal. There doesn’t appear to be a significant difference between the mean drying time for paint type A and the mean drying time for paint type B. The modification does not seem to be effective in reducing drying times.The confidence interval includes only positive values which suggests that the mean drying time for paint type A is greater than the mean drying time for paint type B. The modification seems to be effective in reducing drying times.5 points Question 19Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (sigma1 = sigma2), so that the standard error of the difference between means is obtained by pooling the sample variances. A researcher was interested in comparing the amount of time spent watching television by women and by men. Independent simple random samples of 14 women and 17 men were selected and each person was asked how many hours he or she had watched television during the previous week. The summary statistics are as follows. Construct a 95% confidence interval for mu1 – mu2, the difference between the mean amount of time spent watching television for women and the mean amount of time spent watching television for men.-8.72 hrs < mu1 – mu2 < -2.28 hrs-8.99 hrs < mu1 – mu2 < -2.01 hrs-8.18 hrs < mu1 – mu2 < -2.82 hrs-8.84 hrs < mu1 – mu2 < -2.16 hrs5 points Question 20The two data sets are dependent. Find to the nearest tenth.35.621.446.344.5