The following null a,nd alternative hypotheses need to be tested:
"H_0:\\mu=32"
"H_1:\\mu<32"
This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.
Based on the information provided, the significance level is "\\alpha=0.05," and the critical value for a left-tailed test is "z_c=-1.645."
The rejection region for this left-tailed test is "R=\\{z:z<-1.645\\}"
The z-statistic is computed as follows:
Since it is observed that "z=-15.6525<-1.645=z_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 32, at the 0.05 significance level.
Using the P-value approach: The p-value is "p=0," and since "p=0<0.05," it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean "\\mu"
is less than 32, at the 0.05 significance level.
The hypothesis that the usual age for marriage among less progressive nations is lower than that of rich countries is true at 95% level of confidence.
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