Question

An investigative reporter for the Chicago Tribune reports that the average spending by American adults during the six-day period after Thanksgiving, spanning Black Friday weekend through Cyber Monday, is $100. The population standard deviation (σ) is known to be $20. To verify the report, 100 randomly sampled American adults were surveyed. The sample showed an average spending of $95. Does this show significant evidence that the population average spending differs from $100? Answer this question using a hypothesis test then construct a 95% confidence interval. Does $100 fall within the interval?

Answer 1

Null hypothesis: H₀: The average spending by American adults during the six-day period after Thanksgiving, spanning Black Friday weekend through Cyber Monday is $100.

Alternate hypothesis: H₁: The average spending by American adults during the six-day period after Thanksgiving, spanning Black Friday weekend through Cyber Monday is differenet from $100.

Appropirate test and level of significance:

As the population standard deviation is known and the sample size is greater than 30, Z-test is the most appropriate test and the level of significance is 5%.

Test statistic:

Observed statistics:

Sample mean

Sample size n = 100

Population standard deviation

Observed test statistic:

Critical values and decision rule:

The test is two sided test and so the upper and lower critical values at 5% level of significance are -1.96 and 1.96 respectively. If the observed test statistic value falls outside this range, it will lead to rejection of null hypothesis.

Conclusion:

The observed test statistic is -2.5 and it falls outside the critical values. So there is significant evidence to conclude that the population average spending differs from $100.

Confidence interval:

The 95% confidence interval for mean is given as

Substitute the relevant values in the above interval.

The value $100 does not fall in this interval. So the rejection of null hypothesis is valid.