Answer:
No
Explanation:
here we are interested to test the mean weight of a bag is less than 5 pound.
So null hypothesis is
![H_0:\mu =5](https://img.qammunity.org/2020/formulas/mathematics/college/ey0d5rlq6uygeip139cn2wrmzsm8krlqrr.png)
and alternative hypothesis is
![H_1:\mu <5](https://img.qammunity.org/2020/formulas/mathematics/college/gz93bvia5u14vwavqu5wx4i3eg7rgpnpsc.png)
Let's do one sample z test.
The given information Sample size = n =100
sample mean
,
standard deviation
![\sigma =0.5](https://img.qammunity.org/2020/formulas/mathematics/college/pnhriqmgh36eldv9pkhe8dkmvd69h093xs.png)
Test statistic:Z=
![\frac{\overline{X}-\mu }{(\sigma )/(√(n))}](https://img.qammunity.org/2020/formulas/mathematics/college/allbqlk8rxvi06sjfnqagxgvi0tck6vp5g.png)
Z=
![(4.38-5)/((0.5)/(√(100)))=-12.4](https://img.qammunity.org/2020/formulas/mathematics/college/2wn9d1lk13u2a70dwvrqfrwb77tp0zfsq0.png)
This is a left tailed test, hence the p value for the left tailed test and Z score= -12.4 is
0
Hence We can reject the null hypothesis to say that we have evidence that apple bag weight at the grocery store is less than 5 pounds and it is not a chance variation