Null hypothesis "H_0:p=0.64"
Alternative hypothesis "H_a:p\\ne0.64"
Test statistic: "z=\\frac{0.52-0.64}{\\sqrt{\\frac{0.64(1-0.64)}{100}}}=-2.5"
P-value: "p=2P(Z<-2.5)=0.012."
Since the P-value is less than 0.1, reject the null hypothesis.
There is a significant evidence that the Customer Reports result does not apply to the manager's own store.
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