Question

VULJHUIV 17 A survey of 700 non-fatal accidents in the United States showed that about 23% involved uninsured drivers. The margin of error is 3 percentage points with a 95% confidence. Does the confidence interval support the claim that at least 25% of non-fatal accidents in the United States involve uninsured drivers. Why? No; the interval does not support this claim because the interval contains values that are above 25% O No; the interval does not support this claim because the interval contains some values that are below 25%. Yes; the interval supports this claim because the interval contains values that are above 25%. Yes; the interval supports this claim because most values in the interval are below 25%. QUESTION 15 A system manager at a large corporation believes that 80% of the emails at his company may be spam. He examines 500 emails received and finds that 382 of them (76.4%) are spam. Test the claim that the percentage of emails at his company that are spam is different than 80%. Perform a hypothesis test, using a significance level of 0.05 and write an interpretation of your results. Do not reject Ho. There is not sufficient evidence to support the claim that the percentage of emails that are spam is different than 80%. Do not reject Ho. There is sufficient evidence to support the claim that the percentage of emails that are spam is different than 80% Reject Ho. There is sufficient evidence to support the claim that the percentage of emails that are spam is different than 80%. Reject Ho. There is not sufficient evidence to support the claim that the percentage of emails that are spam is different than 80%.

Answer 1

Answer 14

Yes; the interval supports the claim because most values in the interval are below 25%

We need to construct the 95% confidence interval for the population proportion. We have been provided with the following information: The sample size is N = 700, and the sample proportion is pˉ​​=0.23, and the significance level is α=0.05 Critical Value Based on the information provided, the significance level is α=0.05, therefore the critical value for this Two-tailed test is Zc​=1.96. This can be found by either using excel or the Z distribution table. Margin of Error The confidence interval: Therefore, based on the data provided, the 95% confidence interval for the population proportion is 0.1988<p<0.2612, which indicates that we are 95% confident that the true population proportion p is contained by the interval (0.1988,0.2612)

Answer 15

Reject H0. there is sufficient evidence to support the claim that the percentage of emails that are spam are different than 80%

The following information is provided: The sample size is N = 500, the number of favorable cases is X = 382 and the sample proportion is pˉ​=X/N​=382/500​=0.764, and the significance level is α=0.05 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p =0.8 Ha: p ≠0.8 This corresponds to a Two-tailed test, for which a z-test for one population proportion needs to be used. (2a) Critical Value Based on the information provided, the significance level is α=0.05, therefore the critical value for this Two-tailed test is Zc​=1.96. This can be found by either using excel or the Z distribution table. (2b) Rejection Region The rejection region for this Two-tailed test is |Z|>1.96 i.e. Z>1.96 or Z<-1.96 (3) Test Statistics The z-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is p =P(|Z|>2.0125)=0.0442 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that |Z|=2.0125 > Zc​=1.96, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.0442, and since p=0.0442≤0.05, it is concluded that the null hypothesis is rejected. (6) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is different than 0.8, at the 0.05 significance level.

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