The president of a university claimed that the entering class this year appeared to be larger than the

The president of a university claimed that the entering class this year appeared to be larger than the entering class from previous years but their mean SAT score is lower than previous years. He took a sample of 20 of this year's entering students and found that their mean SAT score is 1,501 with a standard deviation of 53. The university's record indicates that the mean SAT score for entering students from previous years is 1,520. He wants to find out if his claim is supported by the evidence at a 5% level of significance. a. state the null and alternative hypotheses for this study. b. what critical value should the president use to determine the rejection region? c. what is the value of the test statistic? d. what is the p value of the test? e. what is your conclusion?

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  1. Critical value = -1.729

    Step by Step explanation:

    Given:

    n = 20

    X' = 1501

    Standard deviation = 53

    Mean, u = 1520

    Level of significance, a = 0.05

    The null and alternative hypotheses:

    H0 : u = 1520

    H1 : u < 1520

    This is a lower tailed test.

    Degrees of freedom, df = 20 - 1 = 19

    For critical value:

    [tex]t critical = - t_a, _d_f[/tex]

    From t table df = 19, one tailed

    [tex]t critical = -t _0._0_5, _1_9 = -1.729[/tex]

    Critical value = -1.729

    Decision: Reject null hypothesis H0, if test statistic Z, is less than critical value.

    Test statistic Z =

    [tex]Z = \frac{X' - u}{\sigma / \sqrt{n}}[/tex]

    [tex]Z = \frac{1501 - 1520}{53/ \sqrt{20}}= -1.603[/tex]

    Z = -1.603

    For p-value:

    From excel,

    P(t< -1.603) = t.dist( -1.603, 19, 1)

    = 0.06269

    ≈ 0.0627

    P value = 0.0627

    Since test statistic Z, -1.603, is greater than critical value, -1.729, we fail to reject the null hypothesis H0.

    There is not enough statistical evidence to conclude that mean is less than 1520.

  2. Step-by-step explanation:

    To test the hypothesis is the mean SAT score is less than 1520 at 5% significance level

    The nul hypothesis is

    [tex]H_0; \mu \geq 1520[/tex]

    The alternative hypothesis is

    [tex]H_0 ; \mu\leq 1520[/tex]

    The test statistic is

    [tex]t=\frac{\bar x- \mu}{(\frac{s}{\sqrt{n} } )}[/tex]

    [tex]t= \frac{1501-1520}{(\frac{53}{\sqrt{20} } )} \\\\=-1.603[/tex]

    The t - test statistics is -1.603

    The t - critical value is,

    The small size is small and left tail test.

    Look in the column headed [tex]\alpha = 0.05[/tex] and the row headed in the t - distribution table by using degree of freedom is,

    d.f = n - 1

    = 20 - 1

    = 19

    The t - critical value is -1.729

    The conclusion is that the t value corresponds to sample statistics is not fall in the critical region, so the null hypothesis is not rejected at 5% level of significance.

    there is insignificance evidence ti indicate that the mean SAT score is less than 1520. The result is not statistically significant

  3. False.

    The null hypothesis failed to be rejected.

    At a significance level of 5%, there is not enough evidence to support the claim that the entering class has a mean SAT score that is significantly lower than 1520.

    Step-by-step explanation:

    This is a hypothesis test for the population mean.

    The claim is that the entering class has a mean SAT score that is significantly lower than 1520.

    Then, the null and alternative hypothesis are:

    [tex]H_0: \mu=1520\\\\H_a:\mu< 1520[/tex]

    The significance level is 0.05.

    The sample has a size n=20.

    The sample mean is M=1501.

    As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=53.

    The estimated standard error of the mean is computed using the formula:

    [tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{53}{\sqrt{20}}=11.851[/tex]

    Then, we can calculate the t-statistic as:

    [tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{1501-1520}{11.851}=\dfrac{-19}{11.851}=-1.6[/tex]

    The degrees of freedom for this sample size are:

    [tex]df=n-1=20-1=19[/tex]

    This test is a left-tailed test, with 19 degrees of freedom and t=-1.6, so the P-value for this test is calculated as (using a t-table):

    [tex]\text{P-value}=P(t[/tex]

    As the P-value (0.063) is bigger than the significance level (0.05), the effect is not significant.

    The null hypothesis failed to be rejected.

    There is not enough evidence to support the claim that the entering class has a mean SAT score that is significantly lower than 1520.

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