Answer to Question #98267 in Statistics and Probability for Riya

Question

The average number of damaged tins per week for a local supermarket is 35 tins. The company wishes to decrease this figure and introduces a training program for the warehouse staff to encourage a more careful approach to handling the goods. After the training program, the number of damaged tins per week is monitored, over a period of 51 weeks, with the following results:
30 44 19 32 25 30 16 41 28 45 26 38 30 33 24 15 48 28 20 18 31 15 32 40 42 29 35 45 12 29 33 23 40 28 20 32 28 26 38 34 22 30 27 36 31 27 37 32 34 22 32
Show that the mean of the above sample is 30 with a standard deviation of 8.35.
Hence, perform a hypothesis test at the 10% significance level and determine if the training program has had the desired effect

Answer 1

Number: $N=51$.

Sum: $\sum=1532$

$$mean=\bar{x}={1532 \over 51}\approx30$$

$$\sum(x_i-\bar{x})^2=3485.9215686275$$

$$Variance=s^2={\sum(x_i-\bar{x})^2 \over N-1}={3485.9215686275 \over 51-1}\approx$$

$$\approx69.718431372549$$

$$standard \ deviation=s=\sqrt{Variance}\approx$$

$$\approx\sqrt{69.718431372549}\approx8.35$$

$H_0:\mu=35$

$H_1:\mu<35$

Left-tail test

The critical value (cutoff point) is $|z|=1.2816$ In left-tail hypothesis testing, any z score less than the critical value will be rejected. 

$$Z={30-35 \over 8.35/\sqrt{51}}\approx-4.2763<-1.2816$$

Since $-4.2763$ is less than $-1.2816,$ we reject the null hypothesis. We accept the alternative hypothesis.

Yes, the training program has had the desired effect.