The amount customers spend is often correlated with the amount of time they spend in stores. According to a national survey of 500,000 shopping trips, the average time customers spend in a clothing store is 30 min. The first, second and third quartiles of the time spent, in min, are also given as follows:

Quartile	Value (min)
$Q_1$	8.5
$Q_2$ (median)	18.5
$Q_3$	36.1

Jack, a clothing store manager, suspects that customers spend less time in his store than the national average. Jack collected a random sample of 50 shopping times. The results, $x$ (in min), are summarised as follows: $\sum x = 1475, \qquad \sum x^2 = 45029$

1. State the symmetry property of a normal distribution

For a normal distribution, the distribution is symmetrical. Therefore, the mean and the median should be equal, and the interquartile differences ($Q_3 - Q_2$ and $Q_2 - Q_1$) should be equal.

2. Compare the data's mean and median or quartiles to show a lack of symmetry

Given the national average (mean) is $30$ min and the median ($Q_2$) is $18.5$ min.
Since Mean $\neq$ Median (or $30 > 18.5$), the distribution is skewed.
Alternatively, comparing quartiles: $Q_3 - Q_2 = 36.1 - 18.5 = 17.6$, and $Q_2 - Q_1 = 18.5 - 8.5 = 10.0$.
Since $17.6 \neq 10.0$, the data is not symmetrical.

1. Determine whether to use a 1-tail or 2-tail test based on Jack's suspicion.

Since Jack suspects that customers spend less time than the national average, he is looking for a decrease in the mean. Therefore, Jack should carry out a 1-tail test.

2. Define the population parameter.

Let $\mu$ be the population mean shopping time, in minutes, for customers in Jack's store.

3. State the null and alternative hypotheses.

The national average is $30$ minutes.
$H_0: \mu = 30$
$H_1: \mu < 30$

1. Calculate the unbiased estimates of the population mean and variance from the sample data.

Unbiased estimate of the population mean:
$\bar{x} = \frac{\sum x}{n} = \frac{1475}{50} = 29.5$

Unbiased estimate of the population variance:
$s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)$
$s^2 = \frac{1}{49} \left( 45029 - \frac{1475^2}{50} \right)$
$s^2 = \frac{1}{49} (45029 - 43512.5) = \frac{1516.5}{49} \approx 30.949$

2. Determine the test statistic and calculate the $p$-value for the lower-tail test.

Under $H_0: \mu = 30$, since $n = 50$ is large, by the Central Limit Theorem the test statistic is approximately standard normal:
$Z = \frac{\bar{x} - \mu}{\sqrt{\frac{s^2}{n}}} = \frac{29.5 - 30}{\sqrt{\frac{1516.5/49}{50}}} \approx -0.6355$

Using a graphing calculator for a lower-tail test ($H_1: \mu < 30$):
$p\text{-value} = \mathrm{P}(Z < -0.6355) \approx 0.2625$

3. State the statistical decision and interpret it in the context of the question.

Since the $p$-value $\approx 0.263$ is greater than the $10\%$ significance level ($\alpha = 0.10$), we do not reject $H_0$.

There is insufficient evidence at the $10\%$ level of significance to conclude that customers spend less time in Jack's store than the national average. Therefore, Jack's suspicion is not valid.

1. State the meaning of the significance level in the specific context of the question.

The 10% significance level is the probability of rejecting the null hypothesis ($H_0: \mu = 30$) when it is in fact true. In context, this means there is a 0.1 (or 10%) probability of concluding that the population mean shopping time in Jack's store is less than 30 minutes, when in reality the mean shopping time is exactly 30 minutes.

1. State the Central Limit Theorem justification based on the sample size

Since the sample size is $n = 50$, which is large ($n \ge 30$), the Central Limit Theorem states that the sample mean $\bar{X}$ follows an approximately normal distribution.

1. Deduce the effect of a larger sum of squares on the sample variance.

The unbiased estimate of the population variance is given by $s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$.
Since $\sum x$ and $n$ remain unchanged, a larger $\sum x^2$ will result in a larger value for $s^2$.

2. Explain the effect on the test statistic and the final conclusion.

The test statistic is calculated using $z = \frac{\bar{x} - \mu}{s/\sqrt{n}}$.
Since the sample mean $\bar{x} = \frac{1475}{50} = 29.5$ is less than the population mean $\mu = 30$, the test statistic $z$ is negative.
A larger $s$ increases the denominator, making the test statistic $z$ less negative (i.e. its magnitude becomes smaller and it moves closer to 0).
Since the initial conclusion was to not reject $H_0$ (the original $z$ was not in the critical region), a less negative $z$-value will also not fall in the critical region. Therefore, the conclusion remains unchanged.