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 17 of this years entering students and finds that their mean iq score is 117, with a standard deviation of 11. the college records indicate that the mean iq score for entering students from previous years is 115. 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 (h_0) and the alternative hypothesis (h_1). (h_0:square) (h_1:square) (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.) (square) (d) find the p - value. (round to three or more decimal places.) (square) (e) can we conclude that the mean iq score of this years class is greater than that of previous years? oyes ono

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is that the mean IQ score of this - year's class is the same as previous years, and the alternative hypothesis $H_1$ is that the mean IQ score of this - year's class is greater than previous years. So, $H_0:\mu = 115$ and $H_1:\mu>115$.

Step2: Determine test - statistic type

Since the population standard deviation $\sigma$ is unknown and we have a sample standard deviation $s$, we use a $t$-test statistic. The formula for the $t$-test statistic in a one - sample case is $t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu_0$ is the hypothesized population mean, $s$ is the sample standard deviation, and $n$ is the sample size.

Step3: Calculate the test - statistic value

We are given that $\bar{x} = 117$, $\mu_0=115$, $s = 11$, and $n = 17$.
\[

$$\begin{align*} t&=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}\\ &=\frac{117 - 115}{11/\sqrt{17}}\\ &=\frac{2}{11/\sqrt{17}}\\ &=\frac{2\sqrt{17}}{11}\\ &\approx\frac{2\times4.123}{11}\\ &\approx 0.750 \end{align*}$$

\]

Step4: Calculate the p - value

The degrees of freedom is $df=n - 1=17-1 = 16$. Since this is a one - tailed test with $t\approx0.750$ and $df = 16$, we use the $t$-distribution table or a calculator. Using a calculator (e.g., in Python: 1 - scipy.stats.t.cdf(0.75, 16)), the p - value is approximately $0.236$.

Step5: Make a decision

We compare the p - value with the significance level $\alpha=0.10$. Since $p = 0.236>0.10$, we fail to reject the null hypothesis.

Answer:

(a) $H_0:\mu = 115$, $H_1:\mu>115$
(b) $t$-test statistic
(c) $0.750$
(d) $0.236$
(e) No