In: Statistics and Probability
Calculate 80%, 90% and 99% confidence intervals for the standard deviation. Include here the sample standard deviation and sample size -30pts-
S=
n=
〖χ²〗_R=(chi-square right tail)=
〖χ²〗_L=(chi-square left tail)=
80% confidence interval:
Interpretation:
〖χ²〗_R= 〖χ²〗_L= 90% confidence interval:
Interpretation:
〖χ²〗_R= 〖χ²〗_L= 99% confidence interval:
Interpretation:
How do these confidence intervals relate to the population standard deviation you calculated in question #1, when the set of data was treated as a population? -5 pts-
Use the sample you chose to perform a meaningful hypothesis test for the mean. Include all the assumptions, identify clearly the null and alternate hypotheses, show the method used to test the hypothesis and write a clear conclusion
x ̅=
S=
n=
Assumptions/Claim: -10 pts-
Null Hypothesis H0: -5 pts-
Alternate Hypothesis H1: -5 pts-
Method to Test the Hypothesis: -5 pts-
Hypothesis conclusion: -10 pts-
Sample is: 19 24 8 21 70 45 4 13 482 4 8 4 16 0 16 52 30 42 4 5 143 7 18 109 22 26 0 4702 38 71 376 17
The 80% confidence interval for standard deviation
Where,
sample size, n | 32 |
Sample standard deviation, s | 828.1209 |
Sample variance | 685784.2 |
The critical value for chi square is obtained from chi square distribution table table, for significance level = 0.20 degree of freedom = n - 1 = 31
Left tailed
Right tailed
The 90% confidence interval for standard deviation
sample size, n | 32 |
Sample standard deviation, s | 828.1209 |
Sample variance | 685784.2 |
The critical value for chi square is obtained from chi square distribution table table, for significance level = 0.20 degree of freedom = n - 1 = 31
Left tailed
Right tailed
The 95% confidence interval for standard deviation
sample size, n | 32 |
Sample standard deviation, s | 828.1209 |
Sample variance | 685784.2 |
The critical value for chi square is obtained from chi square distribution table table, for significance level = 0.20 degree of freedom = n - 1 = 31
Left tailed
Right tailed
T test
the hypothesis test is performed in following steps
the null hypothesis and the alternate hypothesis
Null hypothesis: Ho: There is no difference between sample and
population ,
Alternate hypothesis: Ha: There is significant difference
between sample and population mean
The single sample t test is used to compare the sample mean to hypothetical population mean The formula for t statistic is,
Population mean is not given let consider
From the data points,
sample mean | 199.875 |
Sample standard deviation, s | 828.1209 |
The level of significance,
The P-value for the t statistic is obtained from t distribution table for t = -0.0166 and significance level = 0.05
Since the P-value is greater than 0.05 at 5% significance level. The null hypothesis is not rejected.There is no significance difference in the sample mean and the population means