Mathematics, 22.04.2020 01:20 jared2461

A 0.05 significance level is used for a hypothesis test of the claim that when parents use a particular method of genderselection, the proportion of baby girls is less than 0.5. Assume that sample data consists of 66 girls in 144 births, so the sample statistic of 11/ 24 EndFraction 11 24 results in a z score that is 1 standard deviation below 0. a) what is the pvalue b) critic value c) area of the critical region

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Answer from: GreenHerbz206

a) P-value=0.16

b) critic value zc=-1.645

c) area of the critical region = 0.05

Step-by-step explanation:

We have a hypothesis test on the population proportion.

The sample proportion is p=66/144=11/24=0.4583.

The z-statistic results in z=-1.

As the alternative hypothesis (the claim) states that the proportion is below 0.5, the test is left-tailed.

The P-value for this type of test and with a z=-1 is calculated as:

P-value=P(z

The critial value depends on the significance level. In this test we have a significance level of 0.05.

The critical value can be looked-up in a standard normal distribution table. Its the value for which the probability of having a statistic below this critical value is equal to the significance level (0.05):

P(z

For this test, the critical value is zc=-1.645.

As the área of the critical region is equal to the probability of having a statistic below this critical value, and this is the significance level, we have a área of the critical region equal to 0.05.

A 0.05 significance level is used for a hypothesis test of the claim that when parents use a particu

Answer from: nagwaelbadawi

a) $z=\frac{0.458 -0.5}{\sqrt{\frac{0.5(1-0.5)}{144}}}=-1.008$

Since is a left tailed test the p value would be:

p_v =P(z

b) $z_{crit}= -1.645$

c) For this case since is a left tailed test the critical region or the rejection zone of the null hypothesis would be:

$(\infty , -1.645)$

Step-by-step explanation:

Data given and notation

n=144 represent the random sample taken

X=66 represent the number of girls

$\hat p=\frac{66}{144}=0.458$ estimated proportion of girls

p_o=0.5 is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportion is less than 0.5.:

Null hypothesis: $p\geq 0.5$

Alternative hypothesis: p < 0.5

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value p_o .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.458 -0.5}{\sqrt{\frac{0.5(1-0.5)}{144}}}=-1.008$

Part a : p value

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

Since is a left tailed test the p value would be:

p_v =P(z

Part b

We want to conduct a left tailed test with $\alpha=0.05$ and we need to find a critical value in the normal standard distribution who accumulates 0.05 of the area in the left and we got:

$z_{crit}= -1.645$

Part c

For this case since is a left tailed test the critical region or the rejection zone of the null hypothesis would be:

$(\infty , -1.645)$

Answer from: Quest

the 1st digit is in the 100s n its 1

step-by-step explanation:

3,589÷18 = 199.389

so the first digit of the quotient is in the 100s n its 1

Answer from: Quest

y-3=-3/2(x+4)

ur very much welcome