Answer: c)
![H_0:\mu= 1.5\\\\H_a:\mu \\eq1.5](https://img.qammunity.org/2020/formulas/mathematics/college/8ttwhu37mtl58zqhfdqsn4unqixpcp0b1v.png)
Explanation:
Given: Nestor Milk Powder is sold in packets with an advertised mean weight of 1.5 kgs.
i.e.
A consumer group wishes to check the accuracy of the advertised mean and takes a sample of 52 packets finding an average weight of 1.49 kgs.
So 1.49 is sample mean but hypothesis is the statement about the parameter which is
.
i.e. he wanted to check whether
or
![\mu \\eq1.5](https://img.qammunity.org/2020/formulas/mathematics/college/vltnq3319cipmzzalhl13h2ccav5lrqgjr.png)
Since null hypothesis
contains equality and alternative hypothesis
is against it.
Therefore, the set of hypotheses that should be used to test the accuracy of advertised weight would be :
![H_0:\mu= 1.5\\\\H_a:\mu \\eq1.5](https://img.qammunity.org/2020/formulas/mathematics/college/8ttwhu37mtl58zqhfdqsn4unqixpcp0b1v.png)
Hence, the correct answer is option c) oH: μ= 1.5; 1H: μ≠1.5