A psychologist is studying smokers' self-images, which she measures by the self-image (SI) score from a personality inventory. For adults in the U.S., the mean SI score from this inventory is about 150. The psychologist gathers a random sample of 16 SI scores of smokers and finds that their mean is 130 and their standard deviation is 31. Assume that the population of SI scores of smokers is normally distributed with mean μ. Based on the sample, can the psychologist conclude that μ is different from 150? Use the 0.05 level of significance.Perform a two-tailed test. Then fill in the table attached.Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. P Progr xO Stude x Week x D Feedt x O McGr xA ALEK XO Laun XC Searc xC Searc xо Оpen Xь МуQ хO Free E Xb Detai XO Find+A www-awn.aleks.com/alekscgi/x/Isl.exe/1o_u-IgNslkr7j8P3jH-IQirVmXK7o_zpyaAMkjWKJh-pmWc8lyVwdR5LM9wuDObwHbzy27qLjCN1ole6wecNjCfVPSOA3qAJtUcw-Okz.Rw.E Apps* BookmarksMY EBAYа АМAZONYAHOOMail - Savage, Tany..MEGA ESSAYSM GMAIL7 Bank of America | H...DEL ONE FCUUSAA | CHECKING...S STUDY POOLOther bookmarksKnowledge CheckQuestion 5TanyaCarry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)pThe null hypothesis:H, :0The alternative hypothesis:H :0D=0OSOThe type of test statistic:(Choose one)O#0O

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Description: A psychologist is studying smokers' self-images, which she measures by the self-image (SI) score from a personality inventory. For adults in the U.S., the mean SI score from this inventory is about 150. The psychologist gathers a random sample of 16

The population of SI scores of smokers is normally distributed with mean μ. The claim of the test is that the μ is different from 150. The hypothesis is, Null hypothesis: H0:μ=150 Alternative hypothesis: H1:μ≠150 A sample of 16 SI scores of smokers with the mean 130 and standard deviation 31. The sample size is not large(less than 30) and the population standard deviation unknown indicating that one sample t test is appropriate test. Thus, the One sample t test is the type of test statistic.The test statistic is, t=x¯-μsn=130-1503116=-207.75=-2.5806 Thus, the test statistic is -2.5806. The degrees of freedom is, df=n-1=16-1=15 The hypothesis is two tailed. The probability of t not equal to 2.5806 with 15 degrees of freedom can be obtained using the excel formula “=T.DIST.2T(2.5806,15)”. The p-value is 0.0209. Thus, the p-value is 0.0209. Decision rule: If p-value≤α, then reject the null hypothesis. Otherwise, do not reject the null hypothesis. Conclusion: The p-value (0.0209) is less than the level of significance (0.05). Based on decision rule, reject the null hypothesis. Yes, at the 0.05 level of significance the psychologist conclude that the mean SI score among smokers is different from 150.

Answered: A psychologist is studying smokers'…

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A psychologist is studying smokers' self-images, which she measures by the self-image (SI) score from a personality inventory. For adults in the U.S., the mean SI score from this inventory is about 150. The psychologist gathers a random sample of 16 SI scores of smokers and finds that their mean is 130 and their standard deviation is 31. Assume that the population of SI scores of smokers is normally distributed with mean μ. Based on the sample, can the psychologist conclude that μ is different from 150? Use the 0.05 level of significance.

Perform a two-tailed test. Then fill in the table attached.

Carry your intermediate computations to at least three decimal places and round your answers as specified in the table.

P Progr x
O Stude x Week x D Feedt x O McGr x
A ALEK X
O Laun X
C Searc x
C Searc x
о Оpen X
ь МуQ х
O Free E X
b Detai X
O Find
+
A www-awn.aleks.com/alekscgi/x/Isl.exe/1o_u-IgNslkr7j8P3jH-IQirVmXK7o_zpyaAMkjWKJh-pmWc8lyVwdR5LM9wuDObwHbzy27qLjCN1ole6wecNjCfVPSOA3qAJtUcw-Okz.
R
w.
E Apps
* Bookmarks
MY EBAY
а АМAZON
YAHOO
Mail - Savage, Tany..
MEGA ESSAYS
M GMAIL
7 Bank of America | H...
DEL ONE FCU
USAA | CHECKING...
S STUDY POOL
Other bookmarks
Knowledge Check
Question 5
Tanya
Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)
p
The null hypothesis:
H, :0
The alternative hypothesis:
H :0
D=0
OSO
The type of test statistic:
(Choose one)
O#0
O<O
The value of the test statistic:
(Round to at least three
decimal places.)
The p-value:
(Round to at least three
decimal places.)
At the 0.05 level of significance, can the
psychologist conclude that the mean SI score
O Yes
O No
among smokers is different from 150?
I Don't Know
Submit
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Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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