normal distribution sample mean

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Kofi owns a cinema. He wishes to increase attendances and so considers offering customers unlimited amounts of free popcorn and soft drinks.

He estimates that the likely increase in attendances would result in his business being more profitable, provided that the mean value of the free items consumed by each customer was less than £1.50.

Before deciding whether to proceed, Kofi offers 60 customers entering the cinema free popcorn and soft drinks. The value of the items consumed by each of these customers has a mean of £1.33 and a standard deviation of £0.45. These customers may be regarded as a random sample of all his current customers.

Examine whether the mean value of free popcorn and soft drinks that would be consumed by his current customers is less than £1.50. Use the 5% significance level:

Unsure what to do here. I think X is normally distributed by (1.5, 0.45^2), but am I unsure? Is the normal distribution instead (1.3, (0.45^2)/60)? And if so how do I do this hypothesis test? For P(X

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In my view it is a hypothesis test. I would recommend a one sided test. The sample mean can be approximately considered as normal distributed since the sample size is larger than $30$ (Thumb rule). The sample mean is approximately distributed as $\overline X\sim \mathcal N\left(1.5, \frac{0.45^2}{60}\right)$.

The hypotheses are

$H_0: \mu\geq 1.5$

$H_1: \mu< 1.5$

The critical value is $\Phi^{-1}(0.05)=-1.645=z_{0.05}$. And the standardized empirical value is $z=\frac{1.33-1.5}{\frac{0.45}{\sqrt{60}}}=-2.93$

Since $z<z_{0.05}$ we reject the null hypothesis

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Let $x$ be the random variable for the value of the items consumed.

Note that $x$ is not necessarily normally distributed (and most likely it probably is not), but it has a mean of $\mu$ and a standard deviation of $\sigma$.

However, by the Central Limit Theorem, $\bar{x}$ is approximately normally distributed with a mean of $\mu$ and a standard deviation of $\sigma / \sqrt{60}$

You can test the hypothesis using $\sigma \approx 0.45$ since the sample size is sufficiently large.