A firm claims that, on average, their workers spend no more than 8 hours a day at the construction site. The amount of time clocked, xx hours, by each worker from a random sample of 150 workers, is summarised as follows. ∑x=1305, ∑x2=12751.\sum x = 1305,\ \sum x^2 = 12751. Test, at the 1% significance level, whether the firm's claim is valid.

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Worked solution

7

1. State the hypotheses.

Let μ\mu be the population mean time clocked by workers, in hours.
H0:μ=8H_0: \mu = 8
H1:μ>8H_1: \mu > 8
Significance level: α=0.01\alpha = 0.01

2. Calculate the sample mean and unbiased estimate of the population variance.

Sample mean, xˉ=1305150=8.7\bar{x} = \frac{1305}{150} = 8.7

Unbiased estimate of population variance, s2=1n−1(∑x2−(∑x)2n)s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)
s2=1149(12751−13052150)=1397.5149=2795298≈9.3792s^2 = \frac{1}{149}\left(12751 - \frac{1305^2}{150}\right) = \frac{1397.5}{149} = \frac{2795}{298} \approx 9.3792

3. Determine the test statistic and compute the $p$-value.

Under H0H_0, since the sample size n=150n = 150 is large, by the Central Limit Theorem, Xˉ\bar{X} is approximately normally distributed.
Test statistic: Z=Xˉ−8s/150≈N(0,1)Z = \frac{\bar{X} - 8}{s / \sqrt{150}} \approx N(0, 1)

Using a graphing calculator (1-tail ZZ-test):
z≈2.799z \approx 2.799
p-value≈0.00256p\text{-value} \approx 0.00256 (to 3 s.f.)

4. State the conclusion in the context of the problem.

Since p-value=0.00256<0.01p\text{-value} = 0.00256 < 0.01, we reject H0H_0.
There is sufficient evidence at the 1%1\% level of significance to conclude that the population mean time clocked is more than 88 hours. Therefore, the firm's claim is not valid.

Answer

Reject H0H_0. The firm's claim is not valid.

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