A random sample of 50 items gives the mean 6.2 and S.D.10.24 .Can it be regarded as drawn from a population with mean 5.4 at 5% level of significance.
1 Answer

(i) The null hypothesis H$_0$ : μ = 5.4 Alternative hypothesis H$_1$ : μ ≠ 5.4

(ii) Calculation of test statistic: Since sample size is large; Z = $\frac{(\bar{X}-μ)}{(\frac{s}{\sqrt{n}})}$ = $\frac{(6.2-5.4)}{\frac{10.24}{\sqrt{50}}}$ = 0.554

(iii) Level of significance : α =0.05

(iv) Critical value : the value of Z$_α$ at 5% level of significance = 1.96

(v) Decision : since the calculated value of |Z| =0.554 is less than the table value Z$_α$=1.96, the null hypothesis is accepted.

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