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andrey2020 [161]
3 years ago
8

What is the interquartile range for this set of data? 11, 19, 35, 42, 60, 72, 80, 85, 88

Mathematics
2 answers:
Fed [463]3 years ago
8 0
Quartiles are 35 and 80.  80 - 35 = 45
AnnZ [28]3 years ago
3 0

Answer:

 

B)  

55.5

Step-by-step explanation:The interquartile range is the third (upper) quartile of a data set minus the first (lower) quartile. Here, the median is 60, dividing the data set into a lower half below the median and an upper half above the median. Each of these halves contains four data values. The upper quartile is the number midway between 80 and 85, or 82.5. The lower quartile is the number midway between 19 and 35, or 27. The interquartile range is 82.5 - 27 = 55.5.

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Ten years ago 53% of American families owned stocks or stock funds. Sample data collected by the Investment Company Institute in
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Answer:

a) The null and alternative hypothesis are:

H_0: \pi=0.53\\\\H_a:\pi

b) If 300 families were sampled, for a significance level of 5%, there is enough evidence to support the claim that a smaller proportion of American families own stocks or stock funds this year than 10 years ago (P-value = 0.001).

 

Step-by-step explanation:

The claim that we want to have evidence to support is that a smaller proportion of American families own stocks or stock funds this year than 10 years ago.

The hypothesis for this test should state:

- For the null hypothesis, that the population proportion is not significantly different from 53%.

H_0:\pi=0.53

- For the alternative hypothesis, that the population proportion is significantly less than 53%.

H_a: \pi

If 300 families are sampled, we can perform a hypothesis test for a proportion.

The claim is that a smaller proportion of American families own stocks or stock funds this year than 10 years ago.

Then, the null and alternative hypothesis are:

H_0: \pi=0.53\\\\H_a:\pi

The significance level is 0.05.

The sample has a size n=300.

The sample proportion is p=0.44.

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.53*0.47}{300}}\\\\\\ \sigma_p=\sqrt{0.00083}=0.029

Then, we can calculate the z-statistic as:

z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.44-0.53+0.5/300}{0.029}=\dfrac{-0.088}{0.029}=-3.065

This test is a left-tailed test, so the P-value for this test is calculated as:

\text{P-value}=P(z

As the P-value (0.001) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that a smaller proportion of American families own stocks or stock funds this year than 10 years ago.

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