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icang [17]
3 years ago
8

According to a Pew Research Center study, in May 2011, 35% of all American adults had a smart phone (one which the user can use

to read email and surf the Internet). A communications professor at a university believes this percentage is higher among community college students.
She selects 300 community college students at random and finds that 120 of them have a smart phone. In testing the hypotheses H0: p = 0.35 versus Ha: p > 0.35, she calculates the test statistic as Z = 1.82. Assume the significance level is ? = 0.10.

Which of the following is an appropriate conclusion for the hypothesis test?

There is enough evidence to show that more than 35% of community college students own a smart phone (P?value = 0.034).

There is enough evidence to show that more than 35% of community college students own a smart phone (P?value = 0.068).

There is not enough evidence to show that more than 35% of community college students own a smart phone (P?value = 0.966).

There is not enough evidence to show that more than 35% of community college students own a smart phone (P?value = 0.034).

Does secondhand smoke increase the risk of a low weight birth? A baby is "low birth weight" if it weighs less than 5.5 pounds at birth. According to the National Center of Health Statistics, about 7.8% of all babies born in the U.S. are categorized as low birth weight. Researchers randomly select 1200 babies whose mothers had extensive exposure to secondhand smoke during pregnancy. 10.4% of the sample are categorized as low birth weight.

Answer the following:

Which of the following are the appropriate null and alternative hypotheses for this research question.

H0: p = 0.078; Ha: p ? 0.078

H0: p = 0.078; Ha: p > 0.078

H0: p = 0.104; Ha: p ? 0.104

H0: ? = 0.104; Ha: ? > 0.104
Mathematics
1 answer:
taurus [48]3 years ago
4 0

Answer:

z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82  

p_v =P(z>1.82)=0.034  

So the p value obtained was a very low value and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis

And the best conclusion would be:

There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

And for the second case the correct system of hypothesis is:

H0: p = 0.078; Ha: p > 0.078

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

\hat p=\frac{120}{300}=0.4 estimated proportion of college students that have a smart phone

p_o=0.35 is the value that we want to test

\alpha=0.1 represent the significance level

Confidence=90% or 0.90

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 the proportion is >0.35.:  

Null hypothesis:p\leq 0.35  

Alternative hypothesis:p > 0.35  

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{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.1. The next step would be calculate the p value for this test.  

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

p_v =P(z>1.82)=0.034  

So the p value obtained was a very low value and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis

And the best conclusion would be:

There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

And for the second case the correct system of hypothesis is:

H0: p = 0.078; Ha: p > 0.078

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7.Sabrina wants to deposit $3,066 into savings accounts at three different banks: Bank of the US,
sweet-ann [11.9K]

The amount Sheila deposits in the Bank of US saving account is $511.

Sabrina has a total of $3,066 she wants to deposit. She has three banks that she wants to deposit her money in.

Let, a represent the amount she would deposit she would invest 7/4 bank.

The amount invested in Catch bank = 6 x a = 6a

The amount invested in Bank of US = 20% x ( 6a + a)

= 0.2 x 7a

= 1.4a

The total amount invested in the three banks can be represented with this equation:

1.4a + a + 6a = 3066

In order to determine the amount she would save in the Bank of US, the amount deposited in 7/4 bank has to be determined first.

8.4a = 3066

a = $365

The amount deposited in the Bank of US = 1.4a

= 1.4 x 365

= $511

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