Since the p value is lower than the significance level given of 0.05 we have enough evidence to reject the null hypothesis on this case. And the best conclusion for this case is:

We (reject) the null hypothesis. That means that we (found) evidence to support the alternative.

Step-by-step explanation:

We have the following info given:

$\bar X = 35.6$ represent the sampel mean for the age of customers

$\sigma = 5$ represent the population standard deviation

$n = 280$ represent the sample size selected

We want to test if the mean age of her customers is over 35 so then the system of hypothesis for this case are:

Null hypothesis: $\mu \leq 35$

Alternative hypothesis $\mu >35$

The statistic for this case is given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing the data given we got:

$z = \frac{35.6-35}{\frac{5}{\sqrt{280}}}= 2.007$

We can calculate the p value since we are conducting a right tailed test like this:

$p_v = P(z>2.007) = 0.0224$

Since the p value is lower than the significance level given of 0.05 we have enough evidence to reject the null hypothesis on this case. And the best conclusion for this case is:

We (reject) the null hypothesis. That means that we (found) evidence to support the alternative.

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3 years ago