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Math  /  Data & Statistics

QuestionGood credit: The Fair Isaac Corporation (FICO) credit score is used by banks and other lenders to determine whether someone is a good credit risk. Scores range from 300 to 850 , with a score of 720 or more indicating that a person is a very good credit risk. An economist wants to determine whether the mean FICO score is lower than the cutoff of 720 . She finds that a random sample of 50 people had a mean FICO score of 707 with a standard deviation of 79 . Can the economist conclude that the mean FICO score is less than 720 ? Use the α=0.10\alpha=0.10 level of significance and the PP-value method with the TI-84 Plus calculator.
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. H0:μ=720H1:μ<720\begin{array}{l} H_{0}: \mu=720 \\ H_{1}: \mu<720 \end{array}
This hypothesis test is a left-tailed ∇\nabla test.
Part 2 of 5 (b) Compute the value of the test statistic. Round the answer to two decimal places. t=−1.16t=-1.16
Part: 2/52 / 5
Part 3 of 5 (c) Compute the PP-value. Round the answer to at least four decimal places. P-value =□P \text {-value }=\square

Studdy Solution

STEP 1

1. The sample size is n=50 n = 50 .
2. The sample mean FICO score is xˉ=707 \bar{x} = 707 .
3. The population standard deviation is s=79 s = 79 .
4. The null hypothesis is H0:μ=720 H_0: \mu = 720 .
5. The alternative hypothesis is H1:μ<720 H_1: \mu < 720 .
6. The level of significance is α=0.10 \alpha = 0.10 .

STEP 2

1. Calculate the test statistic.
2. Find the P-value using the test statistic.

STEP 3

Calculate the test statistic using the formula for the t-statistic:
t=xˉ−μs/nt = \frac{\bar{x} - \mu}{s / \sqrt{n}}
Substitute the given values:
t=707−72079/50t = \frac{707 - 720}{79 / \sqrt{50}}
Calculate the denominator:
s/n=7950≈11.17s / \sqrt{n} = \frac{79}{\sqrt{50}} \approx 11.17
Calculate the test statistic:
t=−1311.17≈−1.16t = \frac{-13}{11.17} \approx -1.16

STEP 4

Use the TI-84 Plus calculator to find the P-value for the calculated test statistic t=−1.16 t = -1.16 with n−1=49 n - 1 = 49 degrees of freedom.
1. Press `2nd` and then `VARS` to access the `DISTR` menu.
2. Select `tcdf(`.
3. Enter the lower bound as −∞-\infty (or a very large negative number), the upper bound as −1.16-1.16, and the degrees of freedom as 4949.
4. Press `ENTER`.
The P-value is approximately:
P-value=0.1265 P \text{-value} = 0.1265
The P-value is:
0.1265 \boxed{0.1265}

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