Answer to Question #250967 in Statistics and Probability for napie

Question

In a survey taken 10 years ago, it was found that 10% of customers of a supermarket brought along their own shopping bags. A recent survey aimed to prove that the current percentage of customers bringing along their own shopping bags is different from 10%. In the survey, it was found that 92 of the 1 000 customers surveyed brought along their own shopping bags. We want to test the claim that the current percentage is not 10%, at the 5% significance level

(a) State the appropriate null and alternative hypothesis.

(b) State and calculate the appropriate test statistic.

(c) Determine the critical value of the test or the p–value of the test.

(d) State whether or not you reject the null hypothesis, giving the reason.

(e) Draw an appropriate conclusion

Answer 1

To answer the question we should test hypothesis about the probability of the event

We got the relative frequency $w = {\frac {92} {100}}$, and want to test if it's significantly different from $p{\scriptscriptstyle 0} = 0.1$

Null hypothesis: p = 0.1

Alternative hypothesis: $p \not =0.1$

Since we want to test the claim that current percenatge is not 10%, then if we reject the null hypothesis it will mean that this claim is correct

The test statistic: $U={\frac {(w-p{\scriptscriptstyle 0})\sqrt{n}} {\sqrt{p{\scriptscriptstyle 0}*(1-p{\scriptscriptstyle 0})}}}$

In our case: $U={\frac {(0.092-0.1)\sqrt{1000}} {\sqrt{0.1*0.9}}}= -0.84$

Due to the form of the alternative hypothesis, two-tailed test is required

The critical interval is $(-\infty;-u{\scriptscriptstyle 0}) ∪(u{\scriptscriptstyle 0};+\infty)$, where $P(N(0,1)>u{\scriptscriptstyle 0}) = {\frac \alpha 2} = 0.025\to u{\scriptscriptstyle 0} = 1.96$

So, the critical interval is $(-\infty;-1.96) ∪(1.96;+\infty)$

Our U-value does not belong to the critical interval, so we fail to reject the null hypothesis, which means there are no statistically significant evidence that current percentage is different from 10%