the mean monthly Internet bill is $32.79 per household. A sample of 50 households in the Belize District showed a sample mean of $30.63. Use a population standard deviation $5.60.
What is the value of the test statistic?
c) What is the p-value?
d) At α = .05, what is your conclusion?
The following null and alternative hypotheses need to be tested:
This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.
b) The z-statistic is computed as follows:
c) The p-value is or
d) Based on the information provided, the significance level is and the critical value for a two-tailed test is
The rejection region for this two-tailed test is
Since it is observed that it is then concluded that the null hypothesis is rejected.
Using the P-value approach:
The p-value is and since it is concluded that the null hypothesis is rejected.
Therefore, there is enough evidence to claim that the population mean is different than 32.79, at the significance level.
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