1.There is discussion in the media of grade inflation; that is, average grades r
ID: 3157682 • Letter: 1
Question
1.There is discussion in the media of grade inflation; that is, average grades rising over time. To determine if grade averages are really higher now than in the past, Professor Smith gathered records from 90 randomly selected students to compare to the mean of 2.54 that existed 10 years ago. The 90 students produced a grade average of 2.69 with a standard deviation of 0.96. At the 0.05 level of significance, are grade averages higher than in the past? (5 step process)
The average hemoglobin reading for a sample of 20 teachers was 16 grams per 100 milliliters, with a sample standard deviation of 2 grams. Find the 99% confidence interval of the true mean. (5 step process)
Explanation / Answer
Answer to first question
Hypothesis:
H0 : Average grade of students is same as that existed 10 years ago.
H1 : Average grade of students is higher than that existed 10 years ago.
Test Statistics
t=
where = grade average of current students, µ = grade average of students 10 years ago, s = standard average of current students, n = sample size(here 90)
We reject null hypothesis if observed t > t value with d.f. 89 and signifincant value 0.05
Now, t follows T distribution with degree of freedom n-1
The reason of this is is assumed to follow Normal distribution (due to central limit theorem) and is assumed to follow Chi square distribution with d.f. n-1.
Now putting values we get
t = (2.69-2.54)/(0.96/90) = 1.482318 (using calculator)
From biometrica table we have t distribution value for d.f. 89 and significant value 0.05 approximately 1.98 which is greater than 1.48(the observed value of test statistics)
Hence at 0.05 level of significance we accept null hypothesis ie we conclude that the grade of students is not higher than that of 10 years ago.
Answer to second question:
We have sample average haemoglobin level of 20 teachers as 16 gms with sample standard variance of 2.
To calculate the 99% confidence interval we use following equation
Interval =
From table = 2.63 (approx)
Hence Upper Limit = 16+(2/20)2.63 =17.17
Lower Limit = 16-(2/20)2.63 = 14.82
Thus confidence interval = (14.82,17.17)
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