The provided sample mean is "\\bar{X}=1000" and the known population standard deviation is "\\sigma=10,"
and the sample size is "n=200."
The following null and alternative hypotheses need to be tested:
"H_0:\\mu \\geq980"
"H_1:\\mu<980"
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=-1.64."
The rejection region for this right-tailed test is "R=\\{z:z<-1.64\\}"
The z-statistic is computed as follows:
Since it is observed that "z=28.284>-1.64=z_c," it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean "\\mu" is less than 980, at the 0.05 significance level. There is not enough evidence to claim that the new technique will be revised.
Using the P-value approach: The p-value is "p=1," and since "p=1>0.05," it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population "\\mu"
is less than 980, at the 0.05 significance level. There is not enough evidence to claim that the new technique will be revised.
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