Question

Please Help with BOTH

1)

2)

Suppose 1 and u2 are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of braking systems. Use the two-sample t test at significance level 0.01 to test Ho: -210 versus H2: H- 42 < -10 for the following data: m 8, x 114.5, s1 5.04, n 8, y 129.2, and s2 = 5.36. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t P-value = State the conclusion in the problem context. OReject Ho. The data suggests that the difference between mean stopping distances is less than -10 OReject Ho. The data does not suggest that the difference between mean stopping distances is less than 10. OFail to reject H. The data suggests that the difference between mean stopping distances is less than 10. OFail to reject H. The data does not suggest that the difference between mean stopping distances is less than -10. You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

Answer Date: 10/11/2019 1) To test the hypothesis is that the difference between population mean stopping distance is less than -10 at 1% significance level. The null and alternative hypothesis is, H:1, - l = -10 H:-!l<-10 The t-test statistics is, By using Excel, find t-test statistics with the help of following steps is: Import the data. Select the MegaStat from Add-Ins option. Select the Compare Two Independent Groups from Hypothesis Tests. Select data input. Select Confidence level, Test difference and alternative hypothesis. Select t-test (pooled variance). Click Ok.

Hypothesis Test: Independent Groups (t-test, pooled variance) Sample 1 Sample 2 114.5 129.2 mean 5.04 5.36 std. dev. 88n 14 df - 14.7000 difference (Sample 1 - Sample 2) 27.0656 pooled variance 5.2025 pooled std. dev. 2.6012 standard error of difference - 10 hypothesized difference -1.81 t .0462 p-value (one-tailed, lower) From the Excel output, the t-test statistics is - 1.81 | The p-value for this test is, From the Excel output, the p-value for this test is 0.046 |.

Decision The conclusion is that the p-value in this context is higher than 0.01 which is 0.046, so the null hypothesis is fail to reject at 1% level of significance. There is insufficient evidence to indicate that the difference between population mean stopping distance is less than -10. The result is not statistically significant. The answer option is 4 - To test the hypothesis is that the true average gap detection threshold for CTS subjects exceeds that for normal subjects at 1% significance level. The null and alternative hypothesis is, H:/ - = 0 H:11-12 <0 The t-test statistics is, By using Excel, find t-test statistics with the help of following steps is: 1) Import the data.

Select the MegaStat from Add-Ins option. Select the Compare Two Independent Groups from Hypothesis Tests. Select data input. Select Confidence level, Test difference and alternative hypothesis. Select t-test (pooled variance). Click Ok. Hypothesis Test: Independent Groups (t-test, pooled variance) Normal 1.91 0.511 CTS 2.56 mean 0.86 std. dev. 10n 16 df -0.65000 difference (Normal - CTS) 0.52982 pooled variance 0.72789 pooled std. dev. 0.34527 standard error of difference 0 hypothesized difference -1.88 t 0390 p-value (one-tailed, lower) From the Excel output, the t-test statistics is -1.88).

The p-value for this test is, From the Excel output, the p-value for this test is 0.039 Decision The conclusion is that the p-value in this context is higher than 0.01 which is 0.039, so the null hypothesis is fail to reject at 1% level of significance. There is insufficient evidence to indicate that the true average gap detection threshold for CTS subjects exceeds that for normal subjects. The result is not statistically significant. The answer option is 1.