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 Mic. Office EXCEL. RM10.00,RM11.50,RM9.70,RM11.20,RM10.70,RM10.20,RM9.70

Question

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 Mic. Office EXCEL.  RM10.00,RM11.50,RM9.70,RM11.20,RM10.70,RM10.20,RM9.70

navratitehf · Accepted Answer

Step 1  Introduction:  Denote μ as the true mean price of sanitizer in District A.  Since it is of interest to know whether the true mean price of sanitizer in District A is RM10.50, the null and alternative hypotheses are:  H0:μ=10.50.  H1:μ≠10.50.  Since the alternative is of the not-equal-to type, the test is two-tailed.  Step 2  Calculation:  The population standard deviation is unknown, and the sample standard deviation must be used for further calculation; further, assume an approximate normal distribution of the sample means. Then, the two-tailed one-sample t-test is to be used.  The sample size is, n=7. The degrees of freedom for a one-sample t-test, for a sample of size n is (n−1). Thus, the degrees of freedom is, n−1=6.  We have entered the data cells A1 to A7 in an Excel worksheet.  In an empty cell, enter =AVERAGE(A1:A7), to obtain the sample average as, x―=10.43.  Again, in an empty cell, enter =STDEV.S(A1:A7), to obtain the sample standard deviation as, s≈0.72045.  The test statistic is: t=x―−μsn  Thus, the test statistic is:  t  t=x―−μsn  =10.43−10.500.720457  =−0.070.720457  ≈−0.257.  Thus, the t-test statistic value is -0.257.  The significance level is, α=0.05.  The p-value for the two-tailed test is:  p-value  =P(|tn−1|≥|t|)  =P(|t6|≥|−0.257|)  =P(|t6|≥|0.257|)  ≈0.8058.  [Using Excel formula: =T.DIST.2T(0.257,6).]  Thus, the p-value is 0.8058.  The p-value is the probability of obtaining a test statistic value at least as extreme as the observed value when the null hypothesis is true.  Decision rule using p-value:  Reject H0 at significance level α, if p-value ≤α. Otherwise, fail to reject H0.  Since p-value (=0.8058)&amp;gt;α(=0.05), fail to reject H0.  Thus, there is not enough evidence to refute the claim that the true mean price of sanitizer in District A is RM10.50, at 0.05 level of significance.

The following are prices of a sanitizer sold at a few selected stores

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