Question

a supermarket has been selling discounted apples in bundles of five at their counters. a random sample of 49 bundles weighs 970 grams on average, with a standard deviation of 70 grams. test the hypothesis that =1000 grams against the alternative hypothesis of < 1000 at 0.06 level of significance.

Answer 1

Hypothesis testing for a mean (σ is unknown, and the variable is normally distributed in the population or n > 30 )

The provided sample mean is $\bar{x}=970$ and the sample standard deviation is $s=70,$ and the size of the sample is $n=49.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=1000$

$H_1:\mu<1000$

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

The number of degrees of freedom are $df=n-1=49-1=48,$ and the significance level is $\alpha=0.06.$ Based on the provided information, the critical t-value for $\alpha=0.06$ and $df=49$ degrees of freedom is $t_c=-1.582951.$

The rejection region for this left-tailed test is $R=\{t:t<-1.582951\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{970-1000}{70/\sqrt{49}}=-3$$

Since it is observed that $t=-3<-1.582951=t_c,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 1000, at the 0.06 significance level.

Use the P-value approach.

The P-value $t=-3, df=49 \alpha=0.06,$ left-tailed, is $p=0.002136.$ Since $p=0.002136<0.06=\alpha,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 1000, at the 0.06 significance level.