The average daily volume of a computer stock in 2011 was $\mu=35.1$ million shares, according to a reliable source. A stock analyst believes that the stock volume in 2018 is different from the 2011 level. Based on a random sample of 30 trading days in 2018 , he finds the sample mean to be 29.8 million shares, with a standard deviation of $\mathrm{s}=15.1$ million shares. Test the hypotheses by constructing a $95 \%$ confidence interval. Complete parts (a) through (c) below.
(c) Will the researcher reject the null hypothesis?
A. Do not reject the null hypothesis because $\mu=35.1$ million shares falls in the confidence interval.
B. Reject the null hypothesis because $\mu=35.1$ million shares does not fall in the confidence interval.
C. Do not reject the null hypothesis because $\mu=35.1$ million shares does not fall in the confidence interval.
D. Reject the null hypothesis because $\mu=35.1$ million shares falls in the confidence interval.

Step 1 ：Given that the average daily volume of a computer stock in 2011 was \(\mu=35.1\) million shares. A stock analyst believes that the stock volume in 2018 is different from the 2011 level. Based on a random sample of 30 trading days in 2018, he finds the sample mean to be 29.8 million shares, with a standard deviation of \(\mathrm{s}=15.1\) million shares. We are to test the hypotheses by constructing a 95% confidence interval.

Step 2 ：The first step is to calculate the confidence interval for the mean of the stock volume in 2018. The formula for a confidence interval is \(\bar{x} \pm z \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the desired confidence level, \(s\) is the sample standard deviation, and \(n\) is the sample size. In this case, \(\bar{x} = 29.8\), \(s = 15.1\), and \(n = 30\). The z-score for a 95% confidence interval is approximately 1.96.

Step 3 ：Substituting the given values into the formula, we get the lower and upper bounds of the confidence interval as approximately 24.40 and 35.20 respectively.

Step 4 ：We then check whether \(\mu = 35.1\) falls within this interval. If it does, we do not reject the null hypothesis; if it does not, we reject the null hypothesis.

Step 5 ：Since the given mean value of 35.1 falls within this interval, we do not reject the null hypothesis.

Step 6 ：Final Answer: \(\boxed{\text{(A) Do not reject the null hypothesis because } \mu=35.1 \text{ million shares falls in the confidence interval.}}\)