Play the piano: Georgianna claims that in a small city renowned for its music school, the average child takes more than 5 years of piano lessons. We have a random sample of 20 children from the city, with a mean of 5.4 years of piano lessons and a standard deviation of 2.2 years. a) Evaluate Georgianna's claim using a hypothesis test. The hypotheses for the test are: Ho: μ-5 Ha: μ.5 Ho: u- 5 На: и> 5 Но: и - 5 На: 5 To execute this hypothesis test we should use a: Z-test T-Test There are 19 degrees of freedom for this test. The test statistic is (round your answer to 3 decimal places): 0.813 The p-value for this hypothesis test is: greater than .25 between .40 and .50 between .20 and .25 less than .20 The conclusion for the hypothesis test is: Since p> a we accept the null hypothesis Since p> a we fail to reject the alternative hypothesis Since p> a we fail to reject the null hypothesis Since p> a we reject the null hypothesis and accept the altermative b) Construct a 95% confidence interval for the number of years students in this city take piano lessons, and interpret it in context of the data. Round your answers to 2 decimal places. We are 95 % confident that the true population mean years of piano lessons taken by the average child in this small town is between years and 2.09 1.73 X year. e 95

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Play the piano: Georgianna claims that in a small city renowned for its music school,
the average child takes more than 5 years of piano lessons. We have a random sample
of 20 children from the city, with a mean of 5.4 years of piano lessons and a standard
deviation of 2.2 years.
a) Evaluate Georgianna's claim using a hypothesis test.
The hypotheses for the test are:
Ho: μ-5
Ha: μ.5
Ho: u- 5
Ha: p> 5
Но: и - 5
Ha: p-5
To execute this hypothesis test we should use a:
Z-test
T-Test
There are 19
degrees of freedom for this test.
The test statistic is (round your answer to 3 decimal places):
0.813
The p-value for this hypothesis test is:
greater than .25
between .40 and .50
between .20 and .25
less than .20
The conclusion for the hypothesis test is:
Since p> a we accept the null hypothesis
Since p> a we fail to reject the alternative hypothesis
Since p> a we fail to reject the null hypothesis
Since p> a we reject the null hypothesis and accept the altermative
b) Construct a 95% confidence interval for the number of years students in this city
take piano lessons, and interpret it in context of the data. Round your answers to 2
decimal places.
We are 95
% confident that the true population mean
years of piano lessons taken by the average child in this small town is between
years and 2.09
1.73
X year.
95
Transcribed Image Text:Play the piano: Georgianna claims that in a small city renowned for its music school, the average child takes more than 5 years of piano lessons. We have a random sample of 20 children from the city, with a mean of 5.4 years of piano lessons and a standard deviation of 2.2 years. a) Evaluate Georgianna's claim using a hypothesis test. The hypotheses for the test are: Ho: μ-5 Ha: μ.5 Ho: u- 5 Ha: p> 5 Но: и - 5 Ha: p-5 To execute this hypothesis test we should use a: Z-test T-Test There are 19 degrees of freedom for this test. The test statistic is (round your answer to 3 decimal places): 0.813 The p-value for this hypothesis test is: greater than .25 between .40 and .50 between .20 and .25 less than .20 The conclusion for the hypothesis test is: Since p> a we accept the null hypothesis Since p> a we fail to reject the alternative hypothesis Since p> a we fail to reject the null hypothesis Since p> a we reject the null hypothesis and accept the altermative b) Construct a 95% confidence interval for the number of years students in this city take piano lessons, and interpret it in context of the data. Round your answers to 2 decimal places. We are 95 % confident that the true population mean years of piano lessons taken by the average child in this small town is between years and 2.09 1.73 X year. 95
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