Math homework help
I CAN CHECK THESE ANSWERS ONCE YOU ARE COMPLETE
4.
| A sample of 49 observations is taken from a normal population with a standard deviation of 10. The sample mean is 55. |
| Determine the 99 percent confidence interval for the population mean. (Round your answers to 3 decimal places.) |
| Confidence interval for the population mean is and . |
5.
| Forty-nine items are randomly selected from a population of 500 items. The sample mean is 40 and the sample standard deviation 9. |
| Develop a 99 percent confidence interval for the population mean? (Round your answers to 3 decimal places.) |
| The confidence interval is between and |
The following information applies to questions 7- 11 displayed below.]
| The following information is available. |
| H0: μ ≤ 17 |
| H1: μ > 17 |
| The sample mean is 18 for a sample of 38. The population standard deviation is 3. |
| Use the .01 significance level. |
7.
| Is this a one- or two-tailed test? |
| One-tailed test | |
| Two-tailed test |
8.
| What is the decision rule? |
| Fail to reject H0 when z > 2.33 | |
| Fail to reject H0 when z ≤ 2.33
|
9.
| What is the value of the test statistic? (Round your answer to 2 decimal places.) |
| Value of the test statistic |
10.
| What is your decision regarding H0? |
| Fail to reject H0 | |
| Reject H0 |
11.
| What is the p-value? (Round your answer to 4 decimal places.) |
| p-value |
| 12. According to a study by the American Pet Food Dealers Association, 61 percent of U.S. households own pets. A report is being prepared for an editorial in the San Francisco Chronicle. As a part of the editorial a random sample of 260 households showed 170 own pets. Does this data disagree with the Pet Food Dealers Association data? Use a 0.05 level of significance. |
| (a) | State the null hypothesis and the alternate hypothesis. (Round your answers to 2 decimal places.) |
| H0: π = | |
| H1: π ≠ | |
| (b) | State the decision rule for 0.05 significance level. (Round your answers to 2 decimal places.) |
| H0 is rejected if z is not between and . |
| (c) | Compute the value of the test statistic. (Round your answer to 2 decimal places.) |
| Value of the test statistic |
| (d) | Does this data disagree with the Pet Food Dealers Association data? Use a 0.05 level of significance. |
| H0 is . There is evidence to show the proportion has changed. |
13.
| The following null and alternate hypotheses are given. |
| H0 : μ ≤ 50 |
| H1 : μ > 50 |
| Suppose the population standard deviation is 10. The probability of a Type I error is set at .01 and the probability of a Type II error at .30. |
| Assume that the population mean shifts from 50 to 55. How large a sample is necessary to meet these requirements? (Round your answer to the nearest whole number.) |
| Sample |
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QUIZ #2
| During a national debate on changes to health care, a cable news service performs an opinion poll of 500 small-business owners. It shows that 65 percent of small-business owners do not approve of the changes. |
| (a) | Develop a 95 percent confidence interval for the proportion opposing health care changes. (Round your answers to 4 decimal places.) |
| Confidence interval for the proportion and . |
| (b) | Do majority approve? |
QUIZ #4
| Warren County Telephone Company claims in its annual report that “the typical customer spends $62 per month on local and long-distance service.” A sample of 12 subscribers revealed the following amounts spent last month. |
| $65 | $67 | $64 | $67 | $60 | $64 | $68 | $64 | $64 | $57 | $54 | $67 |
| (a) | What is the point estimate of the population mean? (Round your answer to 3 decimal places. Omit the “$” sign in your response.) |
| Estimated population mean | $ |
| (b) | Develop a 95 percent confidence interval for the population mean. (Round your answers to 3 decimal places. Omit the “$” sign in your response.) |
| Confidence interval for the population mean | $ and $ |
| (c) | Is the company’s claim that the “typical customer” spends $62 per month reasonable? |
QUIZ #5
| A sample of 45 observations is selected from a normal population. The sample mean is 27, and the population standard deviation is 6. Conduct the following test of hypothesis using the .05 significance level. |
| H0 : μ ≤ 26 |
| H1 : μ > 26 |
| (a) | Is this a one- or two-tailed test? |
| (b) | What is the decision rule? (Round your answer to 2 decimal places.) |
| H0 and H1 when z > |
| (c) | What is the value of the test statistic? (Round your answer to 2 decimal places.) |
| Value of the test statistic |
| (d) | What is your decision regarding H0? |
| There is evidence to conclude that the population mean is greater than 26. |
| (e) | What is the p-value? (Round your answer to 4 decimal places.) |
QUIZ #6
| Tina Dennis is the comptroller for Meek Industries. She believes that the current cash flow problem at Meek is due to the slow collection of accounts receivable. She believes that more than 60 percent of the accounts are in arrears more than three months. A random sample of 200 accounts showed that 140 were more than three months old. At the .01 significance level, can she conclude that more than 60 percent of the accounts are in arrears for more than three months? |
| (a) | State the null hypothesis and the alternate hypothesis. (Round your answers to 2 decimal places.) |
| H0: π ≤ | |
| H1: π > | |
| (b) | State the decision rule for .01 significance level. (Round your answer to 2 decimal places.) |
| Reject H0 if z is > |
| (c) | Compute the value of the test statistic. (Round your answer to 2 decimal places.) |
| z value is |
| (d) | At the .01 significance level, can she conclude that more than 60 percent of the accounts are in arrears for more than three months? |
| H0. Ms. Dennis is in concluding that more than 60 percent of the accounts are in arrears for more than three months. |
[The following information applies to the questions 10 – 14 displayed below.]
| The waiting time for customers at MacBurger Restaurants follows a normal distribution with a mean of 3 minutes and a standard deviation of 1 minute. At the Warren Road MacBurger, the quality-assurance department sampled 50 customers and found that the mean waiting time was 2.75 minutes. At the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes?
QUIZ #10.
QUIZ #11.
QUIZ #12.
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QUIZ #13.
| What is your decision regarding H0? |
| Do not reject H0 | ||||
| Reject H0
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#14
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