A=1/2b and 3b:4c, find a:c<br> Can anyone help?

While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who en

AVprozaik [17]

Answer:

$z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}$ (1)

$z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057$

$p_v =P(Z>0.0057)=0.4977$

The p value is a very high value and using the significance level $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.

Step-by-step explanation:

1) Data given and notation

$X_{M}=23$ represent the number of men that said they enjoyed the activity of Saturday afternoon shopping

$X_{W}=8$ represent the number of women that said they enjoyed the activity of Saturday afternoon shopping

$n_{M}=66$ sample of male selected

$n_{W}=23$ sample of demale selected

$p_{M}=\frac{23}{66}=0.34848$ represent the proportion of men that said they enjoyed the activity of Saturday afternoon shopping

$p_{W}=\frac{8}{23}=0.34782$ represent the proportion of women with red/green color blindness

z would represent the statistic (variable of interest)

$p_v$ represent the value for the test (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment , the system of hypothesis would be:

Null hypothesis: $p_{M} \leq p_{W}$

Alternative hypothesis: $p_{M} > p_{W}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}$ (1)

Where $\hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{23+8}{66+23}=0.34831$

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057$

4) Statistical decision

Using the significance level provided $\alpha=0.05$ , the next step would be calculate the p value for this test.

Since is a one side right tail test the p value would be:

$p_v =P(Z>0.0057)=0.4977$

So the p value is a very high value and using the significance level $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.

4 0

3 years ago

Bruce flipped a penny 5 times. It landed on heads 2 times and tails 3 times. If he flips the coin a 6th time, which of the follo

Brums [2.3K]

Answer:

C

Step-by-step explanation:

because if the pattern were 2 on heads and three on tails the next flip would be heads

5 0

3 years ago

Three times the square of a certain positive number exceeds six times the number by nine. find the number:

d1i1m1o1n [39]

With the number being x, we have 3x^2>6(x/9). Expanding, we get 3x^2>6x/9 and 3x^2>2x/3. Dividing by x, we get 3x>2/3 and dividing by 3 we get x>2/9

4 0

3 years ago

7/9 - 2/3 and 2/3 - 1/6 <br>

arsen [322]

Answer:

The answer is 1/9 and 1/2

7 0

3 years ago

Read 2 more answers

An employee was randomly chosen from a company. If the expected value of the age of the employee is _____ years old, the mean ag

Vinil7 [7]

This employee who was randomly chosen represents the population of the company. This means that, any parameter measured through this employee can be considered as the mean or average value for the whole population (the company).

Based on this, taking the age of the employee as the desired parameter, the age of this employee will be considered as the average or mean age for the whole company. This means that the age of the employee and the mean age of all employees are equal.

Therefore, the answer is C) 32, 32
<span>An employee was randomly chosen from a company. If the expected value of the age of the employee is 32 years old, the mean age of all the employees in the company is 32 years old.</span>

5 0

3 years ago

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A=1/2b and 3b:4c, find a:c Can anyone help?