Q:

Paul wants to compare whether the prices of tangerines in two supermarkets differ significantly. He has been noting down the daily prices for the last few days and has found the difference in the sample means to be 20 cents. If the standard deviation of the distribution of the difference between sample means is 28 cents and we are testing the null hypothesis at the 95% confidence level, which statement is true? The difference between the two means is significant at the 95% confidence level, so the null hypothesis must be rejected. The difference between the two means is not significant at the 95% confidence level, so the alternative hypothesis must be rejected. The difference between the two means is not significant at the 95% confidence level, so the null hypothesis must be rejected. The difference between the two means is not significant at the 95% confidence level, so the alternative hypothesis must be accepted. The difference between the two means is significant at the 95% confidence level, so the alternative hypothesis must be rejected.

Accepted Solution

A:
If we have a mean difference of 20 cents, and a standard deviation of 28 cents, this gives us a z-value of 20/28 = 0.7143.
Based on a z-table, the critical z-value for a two-tailed 95% confidence interval is +/-1.96.
Since the z-value is within the bounds: -1.96 < 0.7143 < 1.96, this means that the different is not significant at the 95% CI, and the alternative hypothesis must be rejected.
This is the 2nd choice:
The difference between the two means is not significant at the 95% confidence level, so the alternative hypothesis must be rejected.