A random sample of 172 marketing students was asked to rate, on a scale from 1 (not important) to 5 (extremely important), health benefits as a job characteristic. The sample mean rating was 3.31, and the sample standard deviation was 0.70. Test at the 1% significance level the null hypothesis that the population mean rating is at most 3.0 against the alternative that it is larger than 3.0.
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A random sample of 172 marketing students was asked to rate, on a scale from 1 (not important) to 5 (extremely important), health benefits as a job characteristic. The sample
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- Fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 42 randomly selected people who train in groups and finds that they run a mean of 47.1 miles per week. Assume that the population standard deviation for group runners is known to be 4.4 miles per week. She also interviews a random sample of 47 people who train on their own and finds that they run a mean of 48.5 miles per week. Assume that the population standard deviation for people who run by themselves is 1.8 miles per week. Test the claim at the 0.01 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.The mean number of eggs per person eaten in the United States is 223. Do college students eat a different number of eggs than the average American? The 51 college students surveyed averaged 237 eggs per person and their standard deviation was 39.3. What can be concluded at the a = 0.01 level of significance? %3D a. For this study, we should use Select an answer b. The null and altemative hypotheses would be: H 2 Select an answer v H 2vSelect an answer v c. The test statistic ? (please show your answer to 3 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.) e. The p-value is | ? v a f. Based on this, we should Select an answer V the null hypothesis. g. Thus, the final conclusion is that... O The data suggest that the populaton mean is significantly different from 223 at a = 0.01, so there is statistically significant evidence to conclude that the population mean number of eggs consumed by college students per year is different from 223. Toctor O The data…You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 20 night students, and the sample mean GPA is 2.94 with a standard deviation of 0.8. You sample 55 day students, and the sample mean GPA is 2.56 with a standard deviation of 0.44. Test the claim using a 1% level of significance. Assume the population standard deviations are unequal and that GPAs are normally distributed. What are the correct hypotheses? Based on the hypotheses, find the following:Test Statistic = Critical Values = ±
- Consider a set of data in which the sample mean is 30.230.2 and the sample standard deviation is 7.87.8. Calculate the z-score given that x=31.4x=31.4. Round your answer to two decimal places.You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 35 night students, and the sample mean GPA is 2.91 with a standard deviation of 0.97. You sample 20 day students, and the sample mean GPA is 2.99 with a standard deviation of 0.82. Test the claim using a 1% level of significance. Assume the population standard deviations are unequal and that GPAS are normally distributed. Give answer to at least 4 decimal places. What are the correct hypotheses? Ho: M₁ H₁: M₁ Test Statistic = Critical Values = = ± H₁₂ Based on the hypotheses, find the following: ≠v H₂ OF (Just enter the positive CV.) The correct decision is to Fail to reject the null hypothesis The correct summary would be: There is not enough evidence to support the claim ✓ of night students is different from the mean GPA of day students. o that the mean GPAYou are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 30 night students, and the sample mean GPA is 2.83 with a standard deviation of 0.94. You sample 55 day students, and the sample mean GPA is 2.97 with a standard deviation of 0.92. Test the claim using a 1% level of significance. Assume the population standard deviations are unequal and that GPAs are normally distributed. Give answer to at least 4 decimal places. What are the correct hypotheses?H0: Select an answer μ σ² μ₂ p μ₁ x̄₂ s² x̄₁ = Select an answer μ x̄₂ σ² μ₂ x̄₁ μ₁ s² p H1: Select an answer σ² μ x̄₁ s² p x̄₂ μ₁ μ₂ ? = < > ≤ ≠ ≥ Select an answer μ x̄₂ x̄₁ μ₂ σ² s² μ₁ p Based on the hypotheses, find the following:Test Statistic = Critical Values = ±± (Just enter the positive CV.)The correct decision is to Select an answer Reject the null hypothesis Fail to reject the null hypothesis The correct summary would be: Select an answer There is not enough…
- The mean ACT score for 43 male high school students is 21.1 and the standard deviation is 5.0. The mean ACT score for 56 female high school students is 20.9 and the standard deviation is 4.7. At the 1% significance level, can you reject the claim that male and female high school students have equal ACT score averages? Be sure to use the p-value to make your conclusion.Do shoppers at the mall spend more money on average the day after Thanksgiving compared to the day after Christmas? The 52 randomly surveyed shoppers on the day after Thanksgiving spent an average of $122. Their standard deviation was $29. The 59 randomly surveyed shoppers on the day after Christmas spent an average of $117. Their standard deviation was $35. What can be concluded at the αα = 0.05 level of significance? For this study, we should use The null and alternative hypotheses would be: H0:H0: H1:H1: The test statistic = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is αα Based on this, we should the null hypothesis.I have a question: You are interested in estimating the the mean weight of the local adult population of female white-tailed deer (doe). From past data, you estimate that the standard deviation of all adult female white-tailed deer in this region to be 21 pounds. What sample size would you need to in order to estimate the mean weight of all female white-tailed deer, with a 94% confidence level, to within 4 pounds of the actual weight?
- The salaries of professional baseball players are heavily skewed right with a mean of $3.2 million and a standard deviation of $2 million. The salaries of professional football players are also heavily skewed right with a mean of $1.9 million and a standard deviation of $1.5 million. A random sample of 40 baseball players’ salaries and 35 football players’ salaries is selected. The mean salary is determined for both samples. Let represent the difference in the mean salaries for baseball and football players. Which of the following represents the shape of the sampling distribution for ? skewed right since the populations are both right skewed skewed right since the differences in salaries cannot be negative approximately Normal since both sample sizes are greater than 30 approximately Normal since the sum of the sample sizes is greater than 30A researcher is conducting a study to determine the mean trough dosage of medication for a population. Assume a previous study was conducted for the same medication and the mean trough dose was found to be 490 mg with a standard deviation of 30 mg. If the researcher wants to be 95% confident the true mean trough dosage of the medication is within 10 mg of the true mean trough dosage, what sample size is needed if the study is predicted to have an 76% retention rate? n = 50 n = 71 n = 47 n = 104
