Question

A random sample of 66 bags of white cheddar popcorn​ weighed, on​ average, 5.43 ounces with a standard deviation of 0.25 ounce.

Answer 1

Answer with explanation:

As per given , we have to test hypothesis :

$H_0:\mu=5.7\\\\ H_a:\mu$ , where $\mu$ = Population mean.

Since the alternative hypothesis is left--tailed , so test  is a left-tailed test.

Sample size : n=66

Sample mean : $\overline{x}=5.43$

sample standard deviation : s= 0.25

Also, population standard deviation is not given , so we will perform a left tailed t-test.

Test statistics : $t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

$t=\dfrac{5.43-5.7}{\dfrac{0.25}{\sqrt{66}}}$

$t=\dfrac{-0.27}{\dfrac{0.25}{8.12403840464}}\approx-8.77$

For significance level 0.05 and degree of freedom 65 (df=n-1), we have

Critical t-value =$t^*=t_{0.01, 65}=-1.669$  [Using student's t-distribution table]

Decision : Since -8.77(calculated t- value)< -1.669(Critical value) , it means it falls under rejection region.

i.e. We reject the null hypothesis.

Conclusion : We have sufficient evidence to support the alternative hypothesis μ < 5.7 ​ounces .