Question

A random sample of 64 bags of white cheddar popcorn weighed, on avarage, 5.23 ounces with a standard deviation of 0.24 ounces. Test the hypothesis that mean = 5.5 ounces against the alternative hypothesis, mean < 5.5 ounces at the 0.05 level of significance.

Answer 1

Null hypothesis: $u=5.5$

Alternative hypothesis:$u<5.5$

$\alpha$ = 0.5

Since the sample is big(>30 elements), it is appropriate to use z-statistic for the testing

According to the form of the alternative hypothesis, left-sided test should be used

In our case: $Z=$ ${\frac {(5.23 - 5.5)*\sqrt{64}} {0.24}}=-9$

Critical value can be found in table of z-statistic, and it is equal to -1.64

Test statistic value is negative and smaller than critical value, so the null hypothesis should be rejected

There is a significant evidence that mean weight is less than 5.5 ounce