Last year, 50% of MNM. Inc. employees were female. It is believed that there has been a reduction in the percentage of females in the company. This year, in a random sample of 400 employees. 180 were female. a. What are the null and the alternative hypotheses?b. At 95% confidence using the critical value approach, determine if there has been a significant reduction in the proportion of females.

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Answer:

We conclude that there has been a significant reduction in the proportion of females.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 400

p = 50% = 0.5

Alpha, α = 0.05

Number of women, x = 118

First, we design the null and the alternate hypothesis  

[tex]H_{0}: p = 0.50\\H_A: p < 0.50[/tex]

This is a one-tailed test.  

Formula:

[tex]\hat{p} = \dfrac{x}{n} = \dfrac{118}{400} = 0.45[/tex]

[tex]z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]

Putting the values, we get,

[tex]z = \displaystyle\frac{0.45-0.50}{\sqrt{\frac{0.50(1-0.50)}{400}}} = -2[/tex]

Now, we calculate the critical value.

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = -1.645[/tex]

Since the calculated z-statistic is less than the critical value, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, there has been a significant reduction in the proportion of females.

z statistic is less than the critical value therefore, the null hypothesis is rejected and the alternate hypothesis is accepted.

Given :

  • Last year, 50% of MNM. Inc. employees were female. It is believed that there has been a reduction in the percentage of females in the company.
  • This year, in a random sample of 400 employees. 180 were female.

Given that the sample size is, n = 400, p = 50% = 0.5, [tex]\alpha[/tex] = 0.05 and the number of women is, x = 180.

Now, the null and the alternate hypothesis is given by:

[tex]\rm H_0 ; p = 0.50[/tex]

[tex]\rm H_1 ; p < 0.50[/tex]

Now, use the below formula to find the value of [tex]z_{critical}[/tex]:

[tex]\widehat{p}=\dfrac{x}{n}=\dfrac{180}{400}=0.45[/tex]

[tex]z = \dfrac{\widehat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}} }[/tex]

Now, put the values of known terms in the above equation.

[tex]z =\dfrac{0.45-0.50}{\sqrt{\dfrac{0.5(1-0.5)}{400}} }[/tex]

z = -2

Now, [tex]z_{critical}[/tex] at 0.05 level of significance = -1.645

z statistic is less than the critical value therefore, the null hypothesis is rejected.

For more information, refer to the link given below:

https://brainly.com/question/11034287