Respuesta :
ANSWER
A. Reject the null hypothesis. There is not enough evidence to oppose the microwave company's claim.
EXPLANATION
The claim is that the microwave can make popcorn in under 2.5 minutes. Hence, the statistical hypotheses are:
[tex]\begin{gathered} H_0\colon\mu\ge2.25 \\ H_1\colon\mu<2.25 \end{gathered}[/tex]where H0 = null hypothesis
H1 = the hypothesis we are interested in proving
The level of significance of the test is not given, hence, we can assume that the level of significance of the test is 5%:
[tex]\alpha=5\%=0.05[/tex]Since the sample size is n = 134, we can use the approximation to the standard normal distribution to calculate the test statistic:
[tex]z=\frac{\bar{x}-\mu}{\frac{S}{\sqrt[]{n}}}[/tex]For the given sample:
[tex]\begin{gathered} n=134 \\ \bar{x}=2.5\text{ min} \\ S=0.25 \\ \mu=2.25\text{ min} \end{gathered}[/tex]Hence, we have that:
[tex]\begin{gathered} z=\frac{2.5-2.25}{\frac{0.25}{\sqrt[]{134}}} \\ z=\frac{0.25}{\frac{0.25}{11.5758}} \\ z=0.25\cdot\frac{11.5758}{0.25} \\ z=11.5758 \end{gathered}[/tex]Now, we have to find the p-value.
If the p-value is greater than the significance level, the decision is to not reject the null hypothesis.
If the p-value is less than the significance level, the decision is to reject the null hypothesis.
For any z-score value greater than 3, the accumulated probability is 0.99999 and for any value below -3, the accumulated probability is 0.
We have that the z-score is 11.5758. This means that:
[tex]\begin{gathered} P(z>11.58)=1-P(z<11.57)=1-0.99999 \\ \Rightarrow P(z>11.58)=0.00001 \end{gathered}[/tex]As we can see, the p-value is less than the significance level, so the decision is to reject the null hypothesis.
This means that the alternative hypothesis:
[tex]H_1\colon\mu<2.25[/tex]is right (or stays) and so, the claim of the company is right.
Hence, the correct answer is option A.
