Question

We want to test the hypothesis that more than 30% of U.S. households have internet access (with a significance level of 5%). We collect a sample of 150 households and find that 57 have access.

Answer #1

Let the proportion of US households having internet access be P.

Then, Sample proportion (p) = 57/150

= 0.38

Null hypothesis (Ho) : P = 0.30

Alternative hypothesis (H1) : P 0.30

Test statistic is given by-

where, po is the value of population proportion under the null hypothesis = 0.30;

n is sample size = 150

Therefore,

z = 2.162

Now, critical value of z at 5% level of significance for the one tailed test is 1.645 as obtained from the z table.

Since, z calculated > critical value of z, we may reject the null hypothesis. So, proportion of US households having internet access is greater than 30%

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Percent of U.S.
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Observed Number
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Married, no children
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Use a 5% level of significance to test the claim that the
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Type of Household
Percent of U.S.
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Observed Number
of Households in
the Community
Married with children
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104
Married, no children
29%
112
Single parent
9%
32
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25%
97
Other (e.g., roommates, siblings)
11%
66
Use a 5% level of significance to test the claim that the
distribution of U.S. households fits the...

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sample of 411 households from a community in Montana are shown
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29%
127
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