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2 answers:
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Answer:
(a) The claim made by the local news anchor is not correct that the average spending during this period in 2009 was $100.
(b) The news anchor’s claim will not be considered reasonable based on a 90% confidence interval.
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for the population mean is,
The factors affecting the width of the interval are:
- Standard deviation.
- Sample size.
- Confidence level.
In this case the 95% confidence interval for the average spending by American adults during the six-day period after Thanksgiving 2009 is provided as ($80.31, $89.11).
On making decisions about a hypothesis test based on confidence interval, the rule followed is:
If the confidence interval contains the null value then the null hypothesis will be accepted or else rejected.
(a)
The hypothesis to test the claim made by the local newspaper anchor is:
<em>H</em>₀: μ = $100 vs. <em>Hₐ: </em>μ <em>≠</em> $100
The 95% confidence interval does not contains the null value, i.e. $100.
Thus, the null hypothesis will be rejected at 5% level of significance.
Conclusion:
The claim made by the local news anchor is not correct that the average spending during this period in 2009 was $100.
(b)
The critical value of the <em>z</em>-statistic increases on increasing the confidence level and decreases on decreasing the confidence level.
So if the confidence level was 90% instead of 95% the critical value of <em>z</em> will be decreased.
And since the critical value is directly proportional to the width of the interval, the width will also decrease on decreasing the critical value.
So the 90% confidence interval will be much narrower than the interval, ($80.31, $89.11).
Hence the null value will still not be included in the interval.
Thus, the news anchor’s claim will not be considered reasonable based on a 90% confidence interval.