The critical value for a two-tailed hypothesis test with a level of significance of α = 0.05 when a standardized test statistic has a standard normal distribution is z* =
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The critical value for a two-tailed hypothesis test with a level of significance of α = 0.05 when a standardized test statistic has a standard normal distribution is z* =
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- The average length of stay of a sample of 20 patients discharged from a general hospital was 7 days with a standard deviation of 2 days. A sample of 22 patients discharged from a chronic disease hospital had an average length of stay of 36 days with a standard deviation of 10 days. Assume that you sampled from normally distributed populations with equal variances. Find the 90% confidence interval for the difference between the average length of hospital stays between these two populations. Use a pooled variance of 54.4. Round the value of your reliability coefficient to three decimal places.A recent national survey found that high school students watched an average (mean) of 7.0 movies per month with a population standard deviation of 0.6. The distribution of number of movies watched per month follows the normal distribution. A random sample of 43 college students revealed that the mean number of movies watched last month was 6.5. At the 0.05 significance level, can we conclude that college students watch fewer movies a month than high school students? State the null hypothesis and the alternate hypothesis. H0: μ ≥ 7.0; H1: μ < 7.0 H0: μ = 7.0; H1: μ ≠ 7.0 H0: μ > 7.0; H1: μ = 7.0 H0: μ ≤ 7.0; H1: μ > 7.0 State the decision rule. Reject H1 if z < –1.645 Reject H0 if z > –1.645 Reject H1 if z > –1.645 Reject H0 if z < –1.645 1. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)A research team conducted a study of VARIABILITY in students' test performance. Researchers assume that individual records are normally distributed with unknown mean and variance. They collected a sample of n = 25 test scores with summaries shown below. Sample Mean = X = 75.2 and Sample SD = S = 15 1. At the significance level, a = 0.01, do researchers have sufficient evidence that the POPULATION VARIANCE would be ABOVE 120? • Show critical value • Evaluate Test Statistic • Formulate Rejection Rule • State your decision 2. At the significance level, a = 0.01, do researchers have sufficient evidence that the POPULATION VARIANCE would be BELOW 540? • Show critical value • Evaluate Test Statistic • Formulate Rejection Rule • State your decision Solution
- A marketing research company desires to know the main consumption of milk per week among males over age 26. They believe that the milk consumption has a mean of 2.2 liters and want to construct a 98% confidence interval with a maximum error of 0.07 liters. Assuming a variance of 1.96 liters,Determine the critical value for a left tailed test of a population mean at the a= 0.01 level of significance based on a sample size of n= 9.Is there any difference in the variability in golf scores for players on a women's professional golf tour and players on a men's professional golf tour? A sample of 20 tournament scores from events in a tour for women showed a standard deviation of 2.4628 strokes, and a sample of 30 tournament scores from events in a tour for men showed a standard deviation of 2.2173. Conduct a hypothesis test for equal population variances to determine if there is any statistically significant difference in the variability of golf scores for male and female professional golfers. Use a = 0.10. State the null and alternative hypotheses. 2 H_: 2 Ho H₂01 H: #02 2 2 02 2 022 202 Ho: 1 H₂ 2 1 02 Find the value of the test statistic. (Round your answer to two decimal places.) Find the p-value. (Round your answer to four decimal places.) p-value = State your conclusion. O Reject Ho. We cannot conclude that there is a difference in the variability of golf scores for male and female professional golfers. Do not…
- A researcher measures human calcitonin (HCT) levels in men with rheumatoid arthritis. The researcher randomly selects 26 men with rheumatoid arthritis and measures the samplemean HCT level of 33.8 pg/mL and a sample standard deviation of 10.4 pg/mL. Assume thatHCT for individuals with rheumatoid arthritis follows a normal distribution. Construct a 92%confidence interval for the mean HCT level in men with rheumatoid arthritis. Be sure to: State the value for α. State whether you should use z or t and find the appropriate value. State the parameter your confidence interval is for. Find the confidence interval (to 2 decimals). Write a verbal interpretation of your confidence interval. (For example: We are xx%confident...)Volunteers who had developed a cold within the previous 24 hours were randomized to take either zinc or placebo lozenges every 2 to 3 hours until their cold symptoms were gone. Twenty-five participants took zinc lozenges, and 23 participants took placebo lozenges. The mean overall duration of symptoms for the zinc lozenge group was 4.5 days, and the standard deviation of overall duration of symptoms was 1.6 days. For the placebo group, the mean overall duration of symptoms was 8 days, and the standard deviation was 1.8 days. (a) Calculate a 95% confidence interval for the mean overall duration of symptoms in a population of individuals like those who used the zinc lozenges. (Round your answers to two decimal places.)to days(b) Calculate a 95% confidence interval for the mean overall duration of symptoms in a population of individuals like those who used the placebo lozenges. (Round your answers to two decimal places.)to days(c) On the basis of the intervals computed in parts (a) and…Is there any difference in the variability in golf scores for players on a women's professional golf tour and players on a men's professional golf tour? A sample of 20 tournament scores from events in a tour for women showed a standard deviation of 2.4638 strokes, and a sample of 30 tournament scores from events in a tour for men showed a standard deviation of 2.2121. Conduct a hypothesis test for equal population variances to determine if there is any statistically significant difference in the variability of golf scores for male and female professional golfers. Use a = 0.10. State the null and alternative hypotheses. 2 O Ho: o 2702 H: 01 Ho: 01 02 キ02 2ァ022 して 2. キ02 2502 2 %3D Find the value of the test statistic. (Round your answer to two decimal places.) Find the p-value. (Round your answer to four decimal places.) p-value %3D State your conclusion. O Reject Ho: We can conclude that there is a difference in the variability of golf scores for male and female professional golfers. O…
- A company that produces top-of-the-line batteries claims that its batteries are good, on average, for more than 65 months. A consumer protection agency tested 25 such batteries to check this claim. It found that the mean life of these 25 batteries is 63.4 months, and the standard deviation is 3 months. If we test the company’s claim at a 5% significance level, what is the p-value? Assume normality. A company that produces top-of-the-line batteries claims that its batteries are good, on average, for more than 65 months. A consumer protection agency tested 25 such batteries to check this claim. It found that the mean life of these 25 batteries is 63.4 months, and the standard deviation is 3 months. If we test the company’s claim at a 5% significance level, what is the p-value? Assume normality. 0.0067 0.0135 0.9962 0.9933What is the probability of randomly selecting a z-score greater than z = -0. 75 from a normal distribution? %3DThe amount of water consumed each week by Bronx residences is normally distributed. The population mean of water consumption is 120.3 gallons with a standard deviation of 10.0 gallons. Test the claim at the 0.10 significance level that the average amount of water consumed is not 125 gallons with a sample size of 100 residences. Do you retain or reject the null hypothesis?