The average length in a sample of 700

Himanshu Sanadhya

Himanshu Sanadhya

Answered question

2022-07-11

The average length in a sample of 700 steel bar manufactured from oneprocess is found to be 250 kg with a standard deviation of 30 kg while the corresponding values in a sample of 300 bars from the other process are 300kg and 40kg. Is there significant difference between the means of two process at 0.01 level of significance?

 

Answer & Explanation

xleb123

xleb123

Skilled2023-06-02Added 181 answers

To determine if there is a significant difference between the means of two processes at a significance level of 0.01, we can perform a two-sample t-test.
Let's denote the sample mean, sample standard deviation, and sample size of the first process as x¯1, s1, and n1, respectively. Similarly, let x¯2, s2, and n2 represent the corresponding values for the second process.
Given that the average length of the steel bars in the first process is x¯1=250 kg, with a standard deviation of s1=30 kg, and a sample size of n1=700. For the second process, we have x¯2=300 kg, s2=40 kg, and n2=300.
The null hypothesis H0 assumes that there is no significant difference between the means of the two processes, while the alternative hypothesis H1 assumes that there is a significant difference between the means. Mathematically, this can be written as:
H0:μ1=μ2
H1:μ1μ2
To test this hypothesis, we can calculate the test statistic using the formula:
t=(x¯1x¯2)(μ1μ2)s12n1+s22n2
where μ1 and μ2 are the population means of the two processes.
Next, we can calculate the critical value for a two-tailed test at a significance level of 0.01. Since the sample sizes are large, we can approximate the critical value using a z-table. For a significance level of 0.01, the critical z-value is approximately 2.58.
Finally, we compare the absolute value of the test statistic with the critical value to make a decision. If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
Let's calculate the test statistic:
t=(250300)(0)302700+402300
Simplifying:
t=50900700+1600300
t=501.2857+1.7778
t3.39
Since the absolute value of the test statistic, 3.39, is greater than the critical value of 2.58, we reject the null hypothesis.
Therefore, based on the given data and a significance level of 0.01, there is a significant difference between the means of the two processes.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in High school probability

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?