Question

Name: Week 7 HSCI 390L: Hypothesis Testing Worksheet 1. It has been reported that the average credit card debt for college seniors is $3262. The student senate at a large university feels that their seniors have a debt much less than this. so they conduct a study of 50 randomly selected seniors and finds that the average debt is $2995, and the population standard deviation is $1100. Let's conduct the test based on an alpha level of 0.05. Complete the 6 steps of hypothesis testing. a. State the research question: b. Identify the null and alternative hypotheses: Ho: c. Specify decision rule: d. Calculate the value of z observed and the corresponding p-value: e. Make a decision regarding your hypotheses: f. Interpret your decision:

Answer 1

a) The average debt of the seniors of that university is much less than the average credit card debt of the college seniors.

b)From the above research question the null hypothesis(H₀) and the alternative hypothesis(H_a) are as follows:
H₀ : $\mu$ = 3262
H_a : $\mu$ < 3262
Here population standard deviation($\sigma$) is given and the sample size is greater than 30, so we can use one sample Z-test.

c) Decision rule:

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

The formula of Z-test statistic is as follows:

$Z=\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n}}}$
Where n = sample size = 50

$\bar X$ = sample mean = 2995
and $\sigma$ = population standard deviation = 1100
Plug these values in the above formulas, we get

For left tailed test, the p-value is as follows:

2995 – 3262 Z= -267 155.5635 = -1.71634 = -1.72

p-value = P(Z < z) = P(Z < -1.72)

From the given Z-table , P(Z < -1.72) = 0.0427

Look the following image:

e) Since p-value = 0.0427 < level of significance = 0.05, so we reject the null hypothesis.

Tobey Table entry for z is the area under the standard normal curve to the left of z. . པ། ཏཀྵ .o .0405 -06 - 07 ch o c3000 .403 soasoa3cio3003 02 .co5 0505 000 .000 to ། 0040004 10 0003 3207 oo07 ,000 . 0 0 005005 005 0010 ,009 g 09 0004 0009009 ། 0003 007 007 -3.༠ ། do13 .0017.00200111 001 བ010 01 -29 0019 186 17016601601501500-0014 -2.8 0260028.0 003030022ca21021020019 27 .0035 ,034 ,0 ,0032.0031.0030co3a280270026 260047 ,00450 043.00410040003903 ,0037 36 -2,5 0062 000.0 .00573055 .0054 cos2 .051 ,0049 004a -2. 032.030.047807500730071.0O69008.0066006 230107014 02 ། ,00 .00960094 0091 .00 0997 0034 -2. 2 0139 0136 .012 ,0129 .012s .0122 .0119 .0116 0113 0110 ,0180174.oig 0166016? .0150150150.0146,013 - 2 0228.02220270212 ,0212,02070202,0197019201e9 0183 .0287 .0281 .02 0268026202560250024 .02390233 -1,4.3590351024 03363290322.031403070301 .029 [=17 | +.0427 ,0418 .0418.0409.0401 .0392030403750367 .0548 ,0537.06 .0516,0505 .049504850475.0465 0455 -1. 5 066 ,0655 ,0643་ .0630 61 .0606 .0594 .05820571 ,0559 -1.4 .08080 793 .0778 .0764 ,0749 .07350 21 .0708 .0694 ,0681 -1.3 ,0968 .0951 .934 .918 .0901 .osgs cs59 .0353 .0338 0323 -1.2 .11511131 .1112 .1093 .10751056 .1038 .1020 .1003 0935 -1.1.1357 .1335,1314 .129212711251 .12301210 .1190 1170 -1.0.1587 .1562 .1539 ,1515 .1492,14691446.1423.1401139 0.s .1841 .1814 .1788 ,1762 .1736 .17། 1 .16851660 .1635 1611 -1902༠་

f) Interpretation: At 5%level of significance there is sufficient evidence to conclude that their seniors have a debt much less

than $3262.