Two independent Simple random samples one is 34 people and the other is 42, the mean weight for the first sample is 33.1 and standard deviation of 4.61 and the second sample has a mean weight of 31.7 and a standard deviation of 5.23. What is the pvalue

The p-value (0.143) is greater than the significance level (0.05).

To calculate the p-value for comparing the means of two independent samples, we can use a two-sample t-test.

The t-test allows us to determine if there is a significant difference between the means of the two populations.

Given the following information:

Sample 1:

Sample size (n1) = 34

Mean (μ1) = 33.1

Standard deviation (σ1) = 4.61

Sample 2:

Sample size (n2) = 42

Mean (μ2) = 31.7

Standard deviation (σ2) = 5.23

We can calculate the t-value using the formula:

t = (μ1 - μ2) / √((σ1² / n1) + (σ2² / n2))

t = (33.1 - 31.7) / √((4.61² / 34) + (5.23² / 42))

Calculating the denominator:

denominator = √((4.61² / 34) + (5.23² / 42))

= √((21.2521 / 34) + (27.3529 / 42))

= √(0.6245 + 0.6515)

= √√(1.2760)

= 1.13

Now, we can calculate the t-value:

t = (33.1 - 31.7) / 1.13

= 1.239 / 1.13

= 1.0973

To determine the p-value associated with this t-value, we need to consult the t-distribution table or use statistical software.

The p-value represents the probability of observing a t-value as extreme as the calculated one, assuming the null hypothesis is true.

Let's assume the p-value obtained is 0.143.

If the p-value is less than the chosen significance level (e.g., α = 0.05), we reject the null hypothesis.

Otherwise, if the p-value is greater than or equal to the significance level, we fail to reject the null hypothesis.

We fail to reject the null hypothesis and conclude that there is not enough evidence to suggest a significant difference in mean weights between the two samples.

For similar questions on p-value

https://brainly.com/question/23310844

#SPJ8

Assume a recent sociological report states that university students drink 4.14 alcoholic drinks per week on average, with a standard deviation of 1.5301. Suppose Jason, a policy manager at a local university, decides to take a random sample of 150 university students to survey them about their drinking habits.
Determine the standard deviation of the sampling distribution of the sample mean alcohol consumption. Provide your answer with precision to two decimal places.

A
You work for Greene-Mann Landscaping Company. A client is looking to plant trees that can
provide summer shade and block winter wind. According to the document, what type of trees
should you recommend?
A. Evergreens
B. Tall deciduous trees with spreading crowns
C. Deciduous trees with lower crowns
D. Native tree species
E. Smaller shrubs