1. A supermarket owner believes that the
mean income of its customers is P50,000
per month. One - hundred customers are
randomly selected and asked of thei r
monthly income. The sample mean is
P48,500 per month and the standard
deviation is P3,200. Is there sufficient
evidence to indicate that the mean
income of the customers of the
superma rket is P50,000 per month? Use
�� = 0.05.
The following null and alternative hypotheses need to be tested:
"H_0:\\mu=50000"
"H_1:\\mu\\not=50000"
This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.
Based on the information provided, the significance level is "\\alpha = 0.05," "df=n-1=99" and the critical value for a two-tailed test is "t_c = 1.984217."
The rejection region for this two-tailed test is "R = \\{t:|t|>1.984217\\}."
The t-statistic is computed as follows:
Since it is observed that "|t|=4.6875>1.984217=t_c," it is then concluded that the null hypothesis is rejected.
Using the P-value approach:
The p-value for two-tailed, "df=99" degrees of freedom, "t=-4.6875" is "p=0.000009," and since "p=0.000009<0.05=\\alpha," it is concluded that the null hypothesis is rejected.
Therefore, there is enough evidence to claim that the population mean "\\mu"
is different than 50000, at the "\\alpha = 0.05" significance level.
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