Please help on this two questions!!
1 answer:
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Answer:
A) -1.40
B) -1.71
C) 0.0872
Step-by-step explanation:
Sample size = n = 25
Sample mean = x = 20.2 mg
Sample standard deviation = s = 3.21 mg
Claim: Sample mean is less than 21.1 mg
So, u = 21.1
Null Hypothesis: u ≥ 21.1 mg
Alternate Hypothesis? u < 21.1 mg
Since we have the sample standard deviation, we will use the t-distribution to answer this question
Part A) t-test statistic
The formula to calculate the t-test statistic is:
Using the values, we get:
Thus, the value of t-test statistic is -1.40
Part B) Critical Value
In order to calculate the critical value we need the value of significance level and the number of degrees of freedom.
Degrees of freedom = df = n -1 = 25 - 1 = 24
Using t-table or t-critical value calculator for significance level 0.05 and 24 df the value comes out to be:
t-critical = -1.71
Since this is a left tailed test, we have considered the negative t-critical value. Rejection region will be the values less than -1.71 i.e. values which lie to left of -1.71
Part C) P-value
In order to calculate the P-value we need the t-score which we calculated in part A and the degrees of freedom.
Using the t-table or p value calculator the p value for t score of -1.40 and df 24 comes out to be:
p-value = 0.0872
Conclusion:
Since the p-value is larger than the significance level, we fail to reject the claim.
Also the test-statistic does not lie in the rejection region, so another hint that we cannot reject the claim.
Thus, no conclusion can be drawn about the effectiveness of the filters using these tests.