What is 2/3 of 99 ?
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$=(a^2-10a)-(b^2+6b) +16$
$=[(a^2-2(5)a+25)-25]-[(b^2+2(3)b+9)-9]+16$
$=(a-5)^2-25-(b+3)^2+9+16$
$=(a-5)^2-(b+3)^2$
Answer:
The proportion of college students who work year-round is 58%.
Step-by-step explanation:
The claim made by the education researcher is that 58% of college students work year-round.
A random sample of 400 college students, 232 say they work year-round.
To test the researcher's claim use a one-proportion <em>z</em>-test.
The hypothesis can be defined as follows:
<em>H</em>₀: The proportion of college students who work year-round is 58%, i.e. <em>p</em> = 0.58.
<em>Hₐ</em>: The proportion of college students who work year-round is 58%, i.e. <em>p</em> ≠ 0.58. C
Compute the sample proportion as follows:
Compute the test statistic value as follows:
The test statistic value is 0.
Decision rule:
If the p-value of the test is less than the significance level then the null hypothesis will be rejected.
Compute the p-value for the two-tailed test as follows:
*Use a z-table for the probability.
The p-value of the test is 1.
The p-value of the test is very large when compared to the significance level.
The null hypothesis will not be rejected.
Thus, it can be concluded that the proportion of college students who work year-round is 58%.