Question

The preodent of a company says that a sales person of the company should make more than 40 calls per week on the average. A sample of 70 sales persons in a week revealed the following information:

# of calls	20-30	30-40	40-50	50-60	60-70
# of sales people	7	12	21	18	12

Does the sample data support management requirement at 1% level of significance?

Answer 1

calls;                  20-30   30-40   40-50   50-60   60-70

sales people;     7            12        21            18        12          

total number of sales people=n=70

mid value of calls x=25    35    45    55    65

sales people is frequency (f) and mid value of calls is x.

f*x=175 ,   420,    945,    990,    780

sum of f*x=3310

mean=sum of f*x/70=47.28571

null hypothesis is, H0=mu>40

alternative hypothesis is, H1=mu<=40

sample size is large,i.e., 70, so we will use z statistics for testing of single mean.

z=(xbar-mu)/(sigma/sqrt(n))

we do not have population standard deviation sigma, so we use sample standard deviation s, now z statistics is-

z=(xbar-mu)/(s/sqrt(n))

s^2=1026.122/69=sum((x-mean)^2)/69

s=3.856338

z=(47.28571-40)/(3.856338/sqrt(70))

z=7.28571/0.4609206

z=15.80687

so calculated z 15.80867 is greater than tabulated value of z at 1% level of significance. So reject the null hypothesis. Sales person should make more than 40 calls per week on the average.