A group of 5 patients treated with a medicine 'A' aeigh 42,3,9,48,60 and 4r kg; a second group of 7 patients from the same hospital treated with medicine 'B' weigh 38,42,56,64,68,69 and 62 kg.Do you agree with the claim that medicine 'B' increases the weight significantly? The value of t at 5% level of significance for 10 d.f.is 2.22281

Well, let me put on my clown nose and give you a humorous response!

Considering the weight of patients treated with medicine 'B', it seems like medicine 'B' not only increases weight but also adds an extra 'B' to it. Perhaps that 'B' stands for "Big-boned" or "Bountiful"! But all joking aside, the t-value you provided indicates that there is a significant difference in weight between the two groups at a 5% level of significance. So, it looks like medicine 'B' does have an effect on weight.

Remember, I'm just a clown bot, so take my response with a pinch of laughter!

To determine if medicine 'B' increases the weight significantly compared to medicine 'A', we can perform a t-test.

The null hypothesis (H0) is that there is no significant difference in weight between the two groups, while the alternative hypothesis (H1) is that there is a significant difference.

First, we need to calculate the mean weight for each group:
- Group 'A': (42 + 3 + 9 + 48 + 60 + 4r) / 5
- Group 'B': (38 + 42 + 56 + 64 + 68 + 69 + 62) / 7

Without knowing the value of 'r', we cannot calculate the mean weight for group 'A' precisely. So let's work with the available information for now.

Next, we need to calculate the standard deviation for each group. I'll assume 'r' is a placeholder and use the given weights.

For group 'A':
- Calculate the sum of the squared differences from the mean: (42 - mean)^2 + (3 - mean)^2 + (9 - mean)^2 + (48 - mean)^2 + (60 - mean)^2
- Divide by (n - 1), where n is the number of data points (5 in this case)
- Take the square root of the result to get the standard deviation

For group 'B':
- Do the same calculations as above, but using the weights from the second group

Next, we can calculate the t-statistic using the formula:
t = (mean1 - mean2) / sqrt((s1^2 / n1) + (s2^2 / n2))
- mean1 = mean weight of group 'A'
- mean2 = mean weight of group 'B'
- s1 = standard deviation of group 'A'
- s2 = standard deviation of group 'B'
- n1 = number of data points in group 'A' (5)
- n2 = number of data points in group 'B' (7)

Finally, we can compare the calculated value of 't' with the critical value for the given significance level (5%) and degrees of freedom (10). If the calculated 't' is greater than the critical value, we reject the null hypothesis and conclude that there is a significant difference in weight between the two groups.

In this case, the given critical value is 2.22281, and we need to compare it with the calculated 't' value.

To determine whether medicine 'B' increases weight significantly in comparison to medicine 'A', we can perform a statistical hypothesis test using the t-test. The given value of t at a 5% level of significance for 10 degrees of freedom (d.f.) is 2.22281.

Here's how we can approach the hypothesis testing process:

Step 1: State the null hypothesis (H0) and alternative hypothesis (Ha):
- Null hypothesis (H0): There is no significant difference in weight between patients treated with medicine 'A' and medicine 'B'.
- Alternative hypothesis (Ha): There is a significant difference in weight between patients treated with medicine 'A' and medicine 'B'.

Step 2: Determine the critical value for the t-test using the significance level:
- In this case, the significance level is 5% (0.05), and the degrees of freedom (d.f.) is 10.
- Looking up the critical value for a two-tailed t-test with 10 degrees of freedom at a 5% level of significance, we can find that the critical value is approximately ±2.228.

Step 3: Calculate the t-statistic:
- To calculate the t-statistic, we need to compute the mean and standard deviation for both groups.

For medicine 'A':
- Mean (μA) = (42 + 3 + 9 + 48 + 60 + 4r) / 6
- Standard Deviation (σA) = √[((42 - μA)² + (3 - μA)² + (9 - μA)² + (48 - μA)² + (60 - μA)² + (4r - μA)²) / 5]

For medicine 'B':
- Mean (μB) = (38 + 42 + 56 + 64 + 68 + 69 + 62) / 7
- Standard Deviation (σB) = √[((38 - μB)² + (42 - μB)² + (56 - μB)² + (64 - μB)² + (68 - μB)² + (69 - μB)² + (62 - μB)²) / 6]

Step 4: Calculate the t-statistic:
- t = (μB - μA) / √[(σA² / nA) + (σB² / nB)]
- where nA is the number of patients in group A (5 in this case)
- and nB is the number of patients in group B (7 in this case)

Step 5: Compare the t-statistic with the critical value:
- Check whether the calculated t-statistic falls within the critical region determined by the critical value.
- If the calculated t-statistic is greater than the positive critical value or less than the negative critical value, we reject the null hypothesis (H0).
- If the calculated t-statistic falls within the critical region, we fail to reject the null hypothesis (H0).

In this case, you provided the value of the t-statistic, which is 2.22281. Comparing this value with the critical value of ±2.228, we find that the calculated t-statistic falls within the critical region. Therefore, we reject the null hypothesis.

In conclusion, based on the given t-statistic value, we can agree with the claim that medicine 'B' increases weight significantly compared to medicine 'A'.