A geometric secquence has: a3= 15 a6= 5/27; find common ratio
2 answers:
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Answer:
95% Confidence interval: (39.43, 61.58)
Step-by-step explanation:
We are given the following in the question:
Sample mean, = $50.50
Sample size, n = 15
Alpha, α = 0.05
Sample standard deviation = 20
95% Confidence interval:
Putting the values, we get,
95% Confidence interval: (39.43, 61.58)
Step 1: Writing Null and Alternate Hypothesis
First we need to write the Null and Alternate Hypothesis for this problem.
Researcher claims that mileage will be increased on addition of Fuel additive. So the null and alternate hypothesis will be:
: μ ≤ 25.1 (Null Hypothesis)
: μ > 25.1 (Alternate Hypothesis)
This is a Right Tailed Test. Since population standard deviation is not know, we will use t-test to check the researchers claim.
Step 2: Finding Test Statistic
Sample Mean = x = 26.8
Standard Deviation = s = 3.9
Sample Size = n = 35
Degrees of Freedom = df = n - 1 = 34
Test statistic(t) is given by:
Step 3: Finding p value
Using t table or calculators find the p value for t=2.579 with 34 degrees of freedom for one tailed test.
P value comes out to be:
p = 0.0072
Step 4: Conclusion
Since the p value is lesser than the significance level of 0.05, we reject the Null Hypothesis.
We have enough evidence to support the researcher's claim that the new additive increases the miles per gallon of the cars.
First we need to write the Null and Alternate Hypothesis for this problem.
Researcher claims that mileage will be increased on addition of Fuel additive. So the null and alternate hypothesis will be:
This is a Right Tailed Test. Since population standard deviation is not know, we will use t-test to check the researchers claim.
Step 2: Finding Test Statistic
Sample Mean = x = 26.8
Standard Deviation = s = 3.9
Sample Size = n = 35
Degrees of Freedom = df = n - 1 = 34
Test statistic(t) is given by:
Step 3: Finding p value
Using t table or calculators find the p value for t=2.579 with 34 degrees of freedom for one tailed test.
P value comes out to be:
p = 0.0072
Step 4: Conclusion
Since the p value is lesser than the significance level of 0.05, we reject the Null Hypothesis.
We have enough evidence to support the researcher's claim that the new additive increases the miles per gallon of the cars.