Answer: c) [tex]H_0:\mu= 1.5\\\\H_a:\mu \neq1.5[/tex]
Step-by-step explanation:
Given: Nestor Milk Powder is sold in packets with an advertised mean weight of 1.5 kgs.
i.e. [tex]\mu= 1.5[/tex]
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 [tex]\mu[/tex].
i.e. he wanted to check whether [tex]\mu= 1.5[/tex] or [tex]\mu \neq1.5[/tex]
Since null hypothesis[tex](H_0)[/tex] contains equality and alternative hypothesis[tex](H_a)[/tex] is against it.
Therefore, the set of hypotheses that should be used to test the accuracy of advertised weight would be :
[tex]H_0:\mu= 1.5\\\\H_a:\mu \neq1.5[/tex]
Hence, the correct answer is option c) oH: μ= 1.5; 1H: μ≠1.5
