(a) At the .01 level of significance,is the true mean greater than 10?

Null Hypothesis: Ho: M = 10 :Mean is 10 or less pages

Alternative Hypothesis: H1: M >10 :Mean is more than 10 pages

Significance level=alpha (a) = 0.01 or 1%

No of tails= 1

This is a 1-tailed test because we are testing that the mean is greater than 10 Since sample size= 35 >= 30

we will use normal distribution

This is a one tailed test because we are testing the area in the right tail Since the sample size is large we will use z distribution

Z at the 0.01 level of significance 1 tailed test = 2.3263

z critical= 2.3263

Decision rule

If the sample z value > z critical, 2.3263 reject Null Hypothesis If the sample z value < z critical , 2.3263 accept Null Hypothesis

Alternatively:

If p value < significance level , 0.01 reject Null Hypothesis If p value > significance level , 0.01 accept Null Hypothesis

Hypothesized Mean=M = 10 pages

Standard deviation =s= 4.45 pages

sample size=n= 35

sx=standard error of mean=s/square root of n= 0.7522 = ( 4.45 /square root of 35) sample mean= 14.44 pages

z=(sample mean-M )/sx= 5.9027 =(14.44-10)/0.7522

Reject Null Hypothesis since z> z critical

The mean is greater than 10 pages

9.56

Null Hypothesis:Ho: p = 50.%

Alternative Hypothesis:H1: p > 50.%

Level of significance=alpha (a) = 0.10 or 10.0%

This is a one tailed test because we are testing the area in the right tail Since the sample size is large we will use z distribution

z critical= 1.2816

Decision rule

If the sample z value > z critical, 1.2816 reject Null Hypothesis If the sample z value < z critical , 1.2816 accept Null Hypothesis

Alternatively:

If p value < significance level , 0.1 reject Null Hypothesis If p value > significance level , 0.1 Accept Null Hypothesis

Decision rule: If the area in the right tails is less than the level of significance (0.10 0r 10%) reject the Null Hypothesis Or alternatively z statistic computed for the sample should be lesser than z corresponding to 10% significance level

p= 50.00%

q= 50.00%

n= 60

sp=standard error of proportion=square root of (pq/n)= 6.4550% =square root of ( 50.% * 50.% / 60)

sample proportion= 63.33% =38/60

z=(sample proportion-population proportion)/ standard error= 2.0651 =(63.33%-50.%)/6.455%

P-value:

This is a one tailed test therefore area in one of the tails is checked Cumulative probability corresponding to the obtained z value of 2.0651 is 98.05% Thus the area in the right tail= 1.95% =100% -98.05%

P-value= 1.95%

For the null hypothesis to hold the area in the right tail should at least be greater than 10.00%

Since this condition is not satisfied accept Alternative Hypothesis:H1: p > 50.% The coin is biased toward heads

9.62

Null Hypothesis:Ho: p = 95.% 95% or less of the orders are processed on the same day Alternative Hypothesis:H1: p > 95.% More than 95% of the orders are processed on the same day Level of significance=alpha (a) = 0.025 or 2.5%

This is a one tailed test because we are testing the area in the right tail Since the sample size is large we will use z distribution

z critical= 1.9600

Decision rule

If the sample z value > z critical, 1.96 reject Null Hypothesis If the sample z value < z critical , 1.96 accept Null Hypothesis

Alternatively:

If p value < significance level , 0.025 reject Null Hypothesis If p value > significance level , 0.025 Accept Null Hypothesis

p= 95.00%

q= 5.00%

n= 500

sp=standard error of proportion=square root of (pq/n)= 0.9750% =square root of ( 95.% * 5.% / 500)

sample proportion= 97.00% =485/500

z=(sample proportion-population proportion)/ standard error= 2.0513 =(97.%-95.%)/.975%

This is a one tailed test therefore area in one of the tails is checked Cumulative probability corresponding to the obtained z value of 2.0513 is...