The nine item of samples had following value 45,47,50,52,48, 47,49,53 \& 51 .Does the mean of nine item differ significantly from population mean of 47.5 ?

To answer this question, we need to conduct a one-sample t-test. Here are the steps:

1. Calculate the sample mean (x̄): Add up all the values and divide by the number of values.

   (45+47+50+52+48+47+49+53+51)/9 = 49

2. Calculate the sample standard deviation (s): Subtract each value from the mean, square the result, add all these, divide by the number of samples minus 1, and then take the square root.

   √[(45-49)²+(47-49)²+(50-49)²+(52-49)²+(48-49)²+(47-49)²+(49-49)²+(53-49)²+(51-49)²]/(9-1) = 2.45

3. Calculate the standard error (SE): Divide the standard deviation by the square root of the number of samples.

   2.45/√9 = 0.82

4. Calculate the t-score: Subtract the population mean from the sample mean and divide by the standard error.

   (49-47.5)/0.82 = 1.83

5. Determine the degrees of freedom (df): Subtract 1 from the number of samples.

   9-1 = 8

6. Look up the critical t-value for your chosen alpha level (usually 0.05) and your degrees of freedom. For df=8 and alpha=0.05 (two-tailed), the critical t-value is approximately ±2.306.

7. Compare your calculated t-score with the critical t-value. If the absolute value of the t-score is greater than the critical t-value, the difference is significant; if it's less, the difference is not significant.

   |1.83| < 2.306, so the difference is not significant.

Therefore, we cannot conclude that the mean of the nine items differs significantly from the population mean of 47.5.