The one-sample t-statistic for a test of Ho: µ = 42 versus H1: µ < 42 based on n = 15 observations has the value t = -1.343, where Ho and H are the null and alternative hypotheses, respectively. The sample size is denoted by n. Use this t-table to determine which two P-values bracket the P-value of the test. Report your answers in decimal form as listed in the t-table.

Use the t-distribution and the sample results to complete the test of the hypotheses. Use a 5% significance level. Assume the results come from a random sample, and if the sample size is small, assume the underlying distribution is relatively normal.Test H0 : μ=15 vs Ha : μ>15 using the sample results x¯=17.2, s=6.4, with n=40. Give the test statistic and the p-value. What is the conclusion?

The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. An article reports the following summary data on intake for a sample of males age 65-74 years: n = 113, x = 12.4, and s = 6.63. Does this data indicate that average daily zinc intake in the population of all males age 65-74 falls below the recommended allowance? (Use a = 0.05.) State the appropriate null and alternative hypotheses. O Ho: H = 15 H: u > 15 O Ho: H = 15 H: u < 15 Ο H: μ15 H:us 15 Ο Η: μ- 15 H: u = 15 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = p-value = State the conclusion in the problem context. O Reject the null hypothesis. There is sufficient evidence that average daily zinc intake falls below 15 mg/day. O Do not reject the null hypothesis. There is sufficient evidence that average daily zinc intake falls below 15 mg/day. O Do not reject the null hypothesis. There…

14% of all Americans suffer from sleep apnea. A researcher suspects that a lower percentage of those who live in the inner city have sleep apnea. Of the 359 people from the inner city surveyed, 43 of them suffered from sleep apnea. What can be concluded at the level of significance of αα = 0.05? For this study, we should use Select an answer z-test for a population proportion t-test for a population mean  The null and alternative hypotheses would be:     Ho: ? μ p  Select an answer = ≠ > <   (please enter a decimal)    H1: ? p μ  Select an answer < > = ≠   (Please enter a decimal) The test statistic ? t z  =  (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 Select an answer reject fail to reject accept  the null hypothesis. Thus, the final conclusion is that ... The data suggest the populaton proportion is significantly smaller than 14% at αα = 0.05, so…

Use the t-distribution and the sample results to complete the test of the hypotheses. Use a 5% significance level. Assume the results come from a random sample, and if the sample size is small, assume the underlying distribution is relatively normal.Test H0 : μ=4 vs Ha : μ≠4 using the sample results x¯=4.8, s=2.3, with n=15.     Give the test statistic and the p-value.   What is the conclusion?

A researcher randomly assigns college freshmen to either of two experimental conditions. Because both groups consist of college freshmen, someone claims that it is appropriate to use a t test for the two related samples. Comments?