(A) Using TI 84 calculator
press stat then tests then 2-sampZTest
enter the data
press calculate, we get
test statistic = 15.27
z critical value are -2.58 and 2.58 for 0.01 significance level
So, we will reject the null hypothesis as the z statistic is outside critical value range
(b) p value = 0.0000
p value is less than significance level of 0.01, which tells us that the result significant and we reject Ho
We can say that there is sufficient evidence to warrant the rejection of claim that the mean widths are equal
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