According to the United States Census Bureau (2011 data), there are approximately 55,500,000 married couple households in the Un
ited State. Every married-couple household is classified into one of three categories:
• both native which means that both spouses were born in the U.S.
• both foreign which means that both spouses were born outside of the U.S.
• one native, one foreign which means that one spouse was born in the U.S. and one was born outside of the U.S..
If 79.5% of married couple households are both native and 13.2% are both foreign, then
(a) what percentage of married couple households are one native, one foreign
7.3
Correct: Your answer is correct.
% married couple households are one native, one foreign
(b) how many married couple households are one native, one foreign?
Incorrect: Your answer is incorrect.
married couple households are one native, one foreign
• both native which means that both spouses were born in the U.S.
• both foreign which means that both spouses were born outside of the U.S.
• one native, one foreign which means that one spouse was born in the U.S. and one was born outside of the U.S..
If 79.5% of married couple households are both native and 13.2% are both foreign, then
(a) what percentage of married couple households are one native, one foreign
7.3
Correct: Your answer is correct.
% married couple households are one native, one foreign
(b) how many married couple households are one native, one foreign?
Incorrect: Your answer is incorrect.
married couple households are one native, one foreign
1 answer:
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Answer:
We conclude that the percentage of blue candies is equal to 29%.
Step-by-step explanation:
We are given that in a random selection of 100 colored candies, 28% of them are blue. The candy company claims that the percentage of blue candies is equal to 29%.
Let p = <u><em>population percentage of blue candies</em></u>
So, Null Hypothesis, : p = 29% {means that the percentage of blue candies is equal to 29%}
Alternate Hypothesis, : p
29% {means that the percentage of blue candies is different from 29%}
The test statistics that will be used here is <u>One-sample z-test for</u> <u>proportions</u>;
T.S. = ~ N(0,1)
where, = sample proportion of blue coloured candies = 28%
n = sample of colored candies = 100
So, <u><em>the test statistics</em></u> =
= -0.22
The value of the z-test statistics is -0.22.
<u>Also, the P-value of the test statistics is given by;</u>
P-value = P(Z < -0.22) = 1 - P(Z 0.22)
= 1 - 0.5871 = 0.4129
Now, at a 0.10 level of significance, the z table gives a critical value of -1.645 and 1.645 for the two-tailed test.
Since the value of our test statistics lies within the range of critical values of z, <u><em>so we insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.
Therefore, we conclude that the percentage of blue candies is equal to 29%.