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3 years ago

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Find the value of the test statistic z using zequalsStartFraction ModifyingAbove p with caret minus p Over StartRoot StartFracti

on pq Over n EndFraction EndRoot EndFraction . Round to three decimal places. The claim is that the proportion of drowning deaths of children attributable to beaches is more than? 0.25, and the sample statistics include n equals 647 drowning deaths of children with? 30% of them attributable to beaches. A. 2.775 B. -2.937 C. 2.937 D. -2.775

1 answer:

il63 [147K]3 years ago

5 0

Answer:

C. 2.937

Step-by-step explanation:

Given the following :

H0 : p = 0.25

H1 : p > 0.25

Number of observations (n) = 647

p' = 30% = 0.3

SD = sqrt[(p(1 - p) / n)]

1 - p = 1 - 0.25 = 0.75

SD = sqrt[(0.25 * 0.75) / 647]

SD = sqrt[0.0002897]

SD = 0.0170234

Z = (p' - p) / SD

Z = (0.3 - 0.25) / 0.0170234

Z = 0.05 / 0.0170234

Z = 2.9371187 = 2.937

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Step-by-step explanation:

The population of the community is given by the following formula:

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In which P(0) is the initial population and r is the growth rate.

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This means that

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Which lets us find r.

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Applying the 5th root to both sides

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Step-by-step explanation:

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I hope it helps you!

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And for this case the p value is higher than the significance level of 0.05 so then we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case is:

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Step-by-step explanation:

For this problem we have the following info given:

$n = 16$ represent the sample size

$\bar X = 18$ represent the sample mean for the drying time

$s= 6$ represent the sample deviation

We want to test the claim that the mean drying time of their product is, at most, 15 minutes, so then the system of hypothesis are:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu >15$

The statistic is given by this formula since we don't know the population deviation:

$t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}$

And replacing we have:

$t = \frac{18-15}{\frac{6}{\sqrt{16}}}= 2$

Now we can find the degrees of freedom given by:

$df = n-1= 16-1=15$

And the p value since we have a right tailed test is given by:

$p_v= P(t_{15}>2) = 0.0639$

And for this case the p value is higher than the significance level of 0.05 so then we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case is:

t = 2, p-value > 5%, fail to reject the null hypothesis. There is not enough evidence to conclude that the mean drying time is greater than 15 minutes.

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3 years ago