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A company that packages salted and unsalted mixed nuts received a complaint that claimed that the company’s salted packages contain more whole cashews than their unsalted packages do. The quality control department investigated the claim by randomly selecting a sample of 45 of each type of package, counting the number of cashews in each package, and finding the mean and standard deviation for both types of packages. Which of the following are the correct null and alternative hypotheses to test the complaint’s claim, where μS is the mean number of cashews per package of salted nuts and μU is the mean number of cashews per package of unsalted nuts?
A. H0 : mu1 - mu2 = 0; Ha : mu1 - mu2 = 0.
B. H0 : mu1 - mu2 = 0; Ha : mu1 - mu2 > 0.
C. H0 : mu1 - mu2 < 0; Ha : mu1 - mu2 > 0.
D. H0 : mu1 - mu2 = 0; Ha : mu1 - mu2 < 0.
E. H0 : mu1 - mu2 > 0; Ha : mu1 - mu2 = 0.

Sagot :

MrRoyalidn

Answer:

[tex]H_0 : \mu_1 - \mu_2 = 0[/tex]

[tex]H_a : \mu_1 - \mu_2 \ne 0[/tex]

Step-by-step explanation:

Let the salted package be represented with 1 and the unsalted, 2.

So:

[tex]\mu_1:[/tex] Mean of salted package

[tex]\mu_2:[/tex] Mean of unsalted package

Considering the given options, the null hypothesis is that which contains =.

So, the null hypothesis is:

[tex]H_0 : \mu_1 - \mu_2 = 0[/tex]

The opposite of = is [tex]\ne[/tex]. So, the alternate hypothesis, is that which contains [tex]\ne[/tex]

So, the alternate hypothesis is:

[tex]H_a : \mu_1 - \mu_2 \ne 0[/tex]

By applying basic meaning of mean and standard deviation we got that correct null and alternative hypotheses to test the complaint’s claim are

(i) [tex]\mu_{1}-\mu_{2} =0[/tex] ; [tex]\mu1 - \mu2 > 0.[/tex]  and (ii) [tex]\mu_{1}-\mu_{2} =0[/tex] ; [tex]\mu1 - \mu2 < 0.[/tex]

What is standard deviation?

Standard deviation is a number used to tell how measurements for a group are spread out from the average (mean), or expected value

Let assume that  salted package are represented by 1 and the unsalted are  represented  by 2 .

So  Mean of salted package is [tex]\mu_{1}[/tex]

and  Mean of unsalted package is [tex]\mu_{2}[/tex]

Now the null hypothesis is that which contains same mean for both types

[tex]\mu_{1}=\mu_{2}\\\\\mu_{1}-\mu_{2}=0[/tex]

And alternative hypotheses is that which contains different mean for both types

[tex]\mu_{1} \neq \mu_{2}\\\\\mu_{1}-\mu_{2} \neq 0[/tex]

Hence two case are possible for  alternative hypotheses

(i)  [tex]\mu1 - \mu2 > 0.[/tex]

(ii) [tex]\mu1 - \mu2 < 0.[/tex]

By applying basic meaning of mean and standard deviation we got that correct null and alternative hypotheses to test the complaint’s claim are

(i) [tex]\mu_{1}-\mu_{2} =0[/tex] ; [tex]\mu1 - \mu2 > 0.[/tex]  and (ii) [tex]\mu_{1}-\mu_{2} =0[/tex] ; [tex]\mu1 - \mu2 < 0.[/tex]

To learn more about standard deviation visit :https://brainly.com/question/12402189

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