Question

S. Would 0/. Coh Eidnce interal btnaue of wic eps invlved in Cheir teshn

Answer 1

(A).

99.94% confidence intervalis wider than 90% confidence interval .

Consider z test

For 90% confidence, $\alpha$=1-.90=0.10,$\alpha$​​​​​​/2 =0.05

Z$\alpha$/2 =1.96

For 99.94% confidence,$\alpha$=1-0.994=0.006, $\alpha$/2=0.003

Z$\alpha$​​​​​​/2=2.748

Confidence interval

X+-Z$\alpha$/2($\sigma$/√n)

Hence,after the calculation we can show that 99.94% confidence interval is more wider than 90% confidence interval.

(B) . Null hypothesis Ho: $\mu$ =$\mu$o

Alternative hypothesis Ha:$\mu$< $\mu$o or $\mu$>$\mu$o (for one tailed test)

Null hypothesis Ho :$\mu$ ,=$\mu$o

Alternative hypothesis Ha :$\mu$$\neq$$\mu$o (for two tailed test).

Test statistic Z=(x- $\mu$o)/($\sigma$/√n)

P value =1-p(z). For one tailed test

P value =2(1-p(z)) for two tailed test.

Reject Ho if p Value <$\alpha$ .

Not reject Ho if p value>$\alpha$.