Question

thanks to an initiative to recruit top students, an administrator at a college claims that this years entering class must have a greater mean iq score than that of entering classes from previous years. the administrator tests a random sample of 21 of this years entering students and finds that their mean iq score is 114, with a standard deviation of 10. the college records indicate that the mean iq score for entering students from previous years is 110.
is there enough evidence to conclude, at the 0.10 level of significance, that the population mean iq score, μ, of this years class is greater than that of previous years? to answer, assume that the iq scores of this years entering class are approximately normally distributed.
perform a one - tailed test. then complete the parts below.
carry your intermediate computations to three or more decimal places. (if necessary, consult a list of formulas.)
(a) state the null hypothesis h0 and the alternative hypothesis h1.
h0 :
h1 :
(b) determine the type of test statistic to use.
select
(c) find the value of the test statistic. (round to three or more decimal places.)

(d) find the critical value. (round to three or more decimal places.)

(e) can we conclude that the mean iq score of this years class is greater than that of previous years?
yes no

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is that the mean IQ score of this - year's class is equal to that of previous years. The alternative hypothesis $H_1$ is that the mean IQ score of this - year's class is greater than that of previous years. So, $H_0:\mu = 110$ and $H_1:\mu>110$.

Step2: Determine test - statistic type

Since the population standard deviation $\sigma$ is unknown and we have a sample standard deviation $s$, and the population is approximately normally distributed, we use a $t$-test statistic.

Step3: Calculate the test - statistic

The formula for the $t$-test statistic in a one - sample case is $t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}$. Here, $\bar{x} = 114$, $\mu_0 = 110$, $s = 10$, and $n = 21$.
\[t=\frac{114 - 110}{10/\sqrt{21}}\approx\frac{4}{10/4.583}\approx\frac{4}{2.182}\approx1.833\]

Step4: Find the critical value

The degrees of freedom is $df=n - 1=21-1 = 20$. For a one - tailed test with $\alpha = 0.10$ and $df = 20$, using the $t$-distribution table or a calculator, the critical value $t_{\alpha,df}=t_{0.10,20}\approx1.325$.

Step5: Make a decision

Since the calculated $t$-value ($1.833$) is greater than the critical value ($1.325$), we reject the null hypothesis.

Answer:

(a) $H_0:\mu = 110$, $H_1:\mu>110$
(b) $t$-test statistic
(c) $1.833$
(d) $1.325$
(e) Yes