Question

Problem 1

The following are prices of a sanitizer sold at a few selected stores in a District A during the COVID-19 pandemic. Based on the sample data, at 0.05 level of significance, can we conclude the average price of the sanitizer in District A is RM10.50? Make your conclusion using the p-value approach, with the aid of Microsoft. Office EXCEL.

RM10.00, RM11.50, RM9.70, RM11.20, RM10.70, RM10.20, RM9.70

Answer 1

For the given data we get

mean = 10.4286

s.d = 0.7204

The provided sample mean is X = 10.4286 and the sample standard deviation is s = 0.7204, and the sample size is n = 7 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: 4 = 10.5 Ha: 41 * 10.5 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided the significance level is a = 0.05, and the critical value for a two- tailed test is te = 2.447 The rejection region for this two-tailed test is R={t: It > 2.447} (3) Test Statistics The t-statistic is computed as follows: 10.4286 - 10.5 0.262 0.7204/v7 X- μo ta

(4. Decision about the null hypothesis Since it is observed that t = 0.262 <te = 2.447, it is then concluded that the nul hypothesis is not rejected Using the P-value approach: The p-value is p=0.8019, and since p = 0.8019 > 0.05, it is concluded that the nuil hypothesis is not rejected.

Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is enough evidence to claim that

the average price of the sanitizer in District A is RM10.50 at 0.05 level of significance,