Answer question- 21,23 mly selects 25 commuten wit h a standard devia this He ra
ID: 3318864 • Letter: A
Question
Answer question- 21,23
mly selects 25 commuten wit h a standard devia this He randomly t S3 mimes Aa0.10, is be rarnib . d as 3.TS. A random The average family size w lon, sample of families in spsxular wchool disniet resulted in the folow 005, does the average family size differ from the national average or Visits A report by the Gallup Poll staied tha ecloked rwoman visits her physician 5,8 tinmes a yer EXCEL Step by Ste researcher randomly selects 20 women and obtaind erage a 58 times a yca A these data ads 4 asle sample of Is a0y Level to co ffers tron $15,0007 The fir the academic year At 005, can it be concluded that the average is 5.8 visits per year? Use the P-value method. s sti 122. Number of Jobs The U.S. Bureau of Labor and Stutis 1.356 tics reported that a person between the ages of 18 and 34 has had an average of 9.2 jobs. To see if this avenage is correct, a researcher selected a random sample el 8 workers between the ages of 18 and 34 and asked ho many different places they had worked. The reslts we eage msvie ticket in ple of 15 movie tickets city indicated that the as dand deviation of $4.86 can it be concluded that al average? At 0.05, can it be concluded that the mean s 92, Use the P-value method. Give one reason why the respondents might not have given the exact number st jobs that they have worked. verage local cell phone 7 minudes. A random an avcrage of dard deviation of 23. Water Consumption The Old Farmer's Almanic water per peron stated that the average consumption ot per day was 123 gallons. To test the hypothesis thl bs igure may no longer be true, a researcher randomy selected 16 people and found that they used on asenge 19 gallons perday and s-53. At 005, is these e concluded that the n average? y of 15 large U.S te time one way is erce executive feels and wants to enough evidence to say that the Old Farmer's Ala figure might no longer be correct? Use the Pa method.Explanation / Answer
21.
Here we have to test,
H0 : mu = 5.8 Vs H1 : mu not= 5.8
where mu is population mean
Assume alpha = level of significance = 0.05
Here sample size is of 20 which is less than 30 and population standard deviation is unknown so we use one sample t-test.
The test statistic follows t-distribution.
The test statistic is,
t = (Xbar - mu) / (s/sqrt(n))
where Xbar is sample mean
s is sample standard deviation
n is sample size
We can do one sample t-test in MINITAB.
steps :
ENTER data into MINITAB sheet --> Stat --> Basic statistics --> 1-Sample t --> Samples in columns : select data column --> Perform hypothesis test --> Hypothesized mean : 5.8--> Options --> Confidence level :95.0 --> Alternative : NOT equal --> ok --> ok
————— 08-12-2017 09:18:54 ————————————————————
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One-Sample T: x
Test of mu = 5.8 vs mu not = 5.8
Variable N Mean StDev SE Mean
x 20 3.850 2.519 0.563
Variable 95.0% CI T P
x ( 2.671, 5.029) -3.46 0.003
Test statistic = -3.46
P-value = 0.003
P-value <alpha
Reject H0 at 5% level of significance
Conclusion : There is sufficient evidence to say that population mean is differ than 5.8