Question

The average score on a nationally administered aptitude test was 76 and the corresponding standard deviation was 8. In order to evaluate a state’s education system, the scores of 100 of the state’s students were randomly selected. These students had an average score of 72. Test at a significance level of 0.05 if there is a significant difference between the state scores and the national scores

Answer 1

Sample size n = 100

Sample mean `bar(x)` = 72

Sample SD S = 8

Population mean µ = 76

Under the Null hypothesis H₀: p = 76

Against the alternative hypothesis H₀: µ ≠ 76  .....(two mail)

Level of significance µ = 0.05

Test statistic:

z = `(bar(x) - "m")/sigma ∼ "N"(0, 1)`

z = `(72 - 76)/((8/sqrt(100))`

= `(-4)/((8/10))`

= `(-4)/0.8`

z = – 5

|z| = 5

Since alternative hypothesis is of two-tailed test we can take |Z| = 5.

∴ Critical value 5% level of significance is  `"Z" > "Z"_(("a")/2)` = 1.96

Inference:

Since the calculated value is less than table value.

i.e `"Z" > "Z"_(("a")/2)` t 5% level of significance the null hypothesis H0 is rejected.

Therefore, we conclude that there is significant difference between the sample mean and population mean µ = 76 and SD σ = 8.