Week 4_ Discussion_ replies

REPLIES:

Benjamin wrote:

We will be using one sample T-test for hypothesis testing. First, we need to calculate mean and standard deviation for the dataset provided. Mean and Standard deviation are calculated using MS Excel.

Following are the steps followed for one sample t-Test:

 

The provided ample means is x̄ =49.13 and the sample standard deviation is s= 1.7433, and the sample size is 𝓃= 20.

1.      Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:μ≥50

Ha: μ≤50

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

2.      Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a left-tailed test is tc=-1.729.

The rejection region for this left-tailed test is R=t:t≺ -1.729

3.      Test Statistics

The t-statistic is computed as follows:

               = -2.232

4. Decision about the null hypothesis

Since it is observed that t = -2.232 ≺tc = -1.729, it is then concluded that the null hypothesis is rejected.

5. Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is less than 50, at the 0.05 significance level.

At 0.05 significance level, there is enough evidence to claim that the bags are being underfilled. Thus, machinery should be recalibrated.

Zaheerah Wrote:

From the excel calculations, the average of the data is 49. 078 and the standard deviation is 1.7756

Null hypothesis (Ho): µ1 ≥ 50

Alternative hypothesis (H1): µ2 < 50

Level of significance = α = 0.05

Sample size (n) = 20

Decision rule 

The tabulated value from the t-statistic table at α = 0.05 and n = 20

 is -1.729

This is a single-tailed test on the left hand therefore the area of rejection is t < - 1.729 

the calculated t-value

= (49.078 – 50) / (1.7756 /√20)

= - 0.922 / 0.397048

= - 2.32214

From the values obtained, the calculated value (- 2.32214) is less than the tabulated value (-1.729). 

At a significance level of α = 0.05, we reject the null hypothesis (µ1 ≥ 50) and fail to reject the alternative hypothesis (µ2 < 50)

From the calculations, we can conclude that the bags had been undefiled. Therefore, the machines should be recalibrated.