Question

The sample data consist of 23 houses from a specific city yielded the average house price $226,460 and the standard deviation of the house price $11,500. Use a significance level 0.01 to test whether the mean house price of the whole city is more than $220,000. Compute the value of the test statistic, and P-value for the specified hypothesis test and state your conclusion. Assume the house prices of this city follows normal distribution. Test statistic: t = 2.69, p-value = 0.00669. Reject the null hypothesis. There is sufficient evidence to conclude that the mean is greater than $220,000. The evidence against the null hypothesis is very strong. Test statistic: t = 2.69, p-value = 0.0036. Reject the null hypothesis. There is sufficient evidence to conclude that the mean is greater than $220,000. The evidence against the null hypothesis is very strong. Test statistic: t = 0.56, P-value = 0.2877. Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean is greater than $220,000. The evidence against the null hypothesis is weak or none. Test statistic: t = 2.69, P-value = 0.9933. Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean is greater than $220,000. The evidence against the null hypothesis is weak or none.

Answer 1

Solution:

Test statistic :

t =( xbar- mu) /sigma/√n = (226460-220000)/(11500/√23) = 2.69

P-value :

Open t-table for one- tail probability . ( Since right-tail test is used)

Look in n=22 row since degrees of freedom is 22. Try to find the value near to 2.69 in that particular row.

We see that 2.69 lies between 2.508 and 2.819 corresponding to significance level alpha= 0.01 and 0.005 respectively.

Therefore our p-value lies between 0.005 and 0.01.

Hence option a is correct.

Pvalue= 0.00669.

Since pvalue < alpha, we reject null hypothesis.

There is sufficient evidence to conclude that mean is greater than 220000. The evidence against null hypothesis is very strong.