The image shows a hypothesis testing problem involving the number of emails received per day by people who send at least one text message per day.

Here's how to solve it step by step:

a) Find the p-value:

• The problem gives:
  • Population mean $\mu = 26.5$
  • Sample mean $\bar{x} = 20.3$
  • Sample size $n = 100$
  • Sample standard deviation $s = 10.2$

1. Formulate Hypotheses:

   • Null hypothesis $H_0$: $\mu = 26.5$ (people who send at least one text message per day receive the same number of emails)
   • Alternative hypothesis $H_1$: $\mu < 26.5$ (people who send at least one text message per day receive fewer emails)

2. Calculate the test statistic (z-score):

   $z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$ Substituting the values:

   $z = \frac{20.3 - 26.5}{\frac{10.2}{\sqrt{100}}} = \frac{-6.2}{1.02} \approx -6.08$

3. Find the p-value: Using a standard normal table or calculator, the p-value corresponding to $z = -6.08$ is extremely small. From the image, the answer is $0.0000001$ (or $1 \times 10^{-7}$).

b) Conclusion at a 0.01 significance level:

Since the p-value $0.0000001$ is much smaller than 0.01, we reject the null hypothesis. Therefore, there is strong evidence to support the claim that people who send at least one text message per day receive fewer emails on average than 26.5 emails per day.

Do you want more details or have any questions about this problem? Here are 5 related questions you can explore:

1. How would the test statistic change if the sample size were larger?
2. What if the sample mean were closer to the population mean? How would that affect the p-value?
3. Could a two-tailed test be applied to this situation? Why or why not?
4. How does the significance level (0.01) influence the hypothesis test?
5. What assumptions do we make when using the z-test for this problem?

Tip:

When performing hypothesis tests, always double-check if the sample size is large enough to use the z-distribution. For small samples, you might need a t-distribution instead.