A Snack Food Company Produces Bags Of Peanuts Labeled As Con ✓ Solved
A Snack Food Company Produces Bags Of Peanuts Labeled As Containing
A snack food company produces bags of peanuts labeled as containing 4 ounces. A consumer reports organization wants to see if the weight is actually less than 4 ounces. They randomly choose 40 bags and their contents are weighed. They find the average weight is 3.5 ounces with a standard deviation of s = 0.9 ounces. Is this sufficient evidence to conclude that the bags contain less than 4 ounces of peanuts?
a. State the null and alternative hypotheses. H0: Ha:
b. What is the value of the one-sample t statistic? Do not pool variances. t = Round to 3 places.
c. What is the P -value for the t test? P -value = Round to 4 places.
d. Is there sufficient evidence that the bags contain less than 4 ounces of peanuts?
e. Give a 95% confidence interval for the mean weight of peanuts in each bag. (The t critical value is 2.009.) From ounces to ounces. Round each number to 2 places.
Paper For Above Instructions
The analysis of the weight of peanut bags produced by a snack food company involves statistical hypothesis testing to determine if the average weight is significantly less than the labeled content of 4 ounces. This study will address the hypotheses, calculate the t-statistic, determine the p-value, assess the evidence against the null hypothesis, and establish a 95% confidence interval.
Hypotheses
The first step in hypothesis testing is to establish the null hypothesis (H0) and the alternative hypothesis (Ha). In this scenario:
- Null Hypothesis (H0): The mean weight of the bags of peanuts is equal to 4 ounces (μ = 4).
- Alternative Hypothesis (Ha): The mean weight of the bags of peanuts is less than 4 ounces (μ
Calculating the t-statistic
To calculate the one-sample t-statistic, the formula used is:
t = (X̄ - μ) / (s / √n)
Where:
- X̄ = sample mean = 3.5 ounces
- μ = population mean under the null hypothesis = 4 ounces
- s = sample standard deviation = 0.9 ounces
- n = sample size = 40
Substituting the values into the formula:
t = (3.5 - 4) / (0.9 / √40)
Calculating the denominator:
0.9 / √40 ≈ 0.142
Now calculating the t-statistic:
t = (-0.5) / 0.142 ≈ -3.521
Rounding to three decimal places, the t-statistic is:
t ≈ -3.521
P-value Calculation
To determine the p-value associated with the t-statistic, we need to refer to the t-distribution with degrees of freedom (df) calculated as:
df = n - 1 = 40 - 1 = 39
The p-value for a one-tailed test can be found using statistical software or a t-distribution table. For t = -3.521 with 39 degrees of freedom, the p-value is found to be approximately:
P-value ≈ 0.0005
Evidence Against the Null Hypothesis
Given the significance level typically set at α = 0.05, we can conclude that if the p-value is less than 0.05, we reject the null hypothesis. Since our calculated p-value of 0.0005 is significantly less than 0.05, we have sufficient evidence to reject H0, thereby concluding that the bags contain less than 4 ounces of peanuts.
Confidence Interval for Mean Weight
To establish a 95% confidence interval for the mean, we use the formula:
CI = X̄ ± (t_critical * (s / √n))
From earlier calculations, we have:
- X̄ = 3.5 ounces
- t_critical = 2.009 (from the given information)
- s = 0.9 ounces
- n = 40
Now calculating the margin of error:
Margin of Error = 2.009 * (0.9 / √40) ≈ 0.284
The confidence interval therefore calculates to:
CI = 3.5 ± 0.284
Rounding to two decimal places, the confidence interval is:
CI: (3.22, 3.78)
Conclusion
The analysis indicates that there is significant evidence to support the claim that the bags of peanuts contain less than the advertised weight of 4 ounces. The one-sample t-test provided a t-statistic of approximately -3.521, with a corresponding p-value of approximately 0.0005. Additionally, the 95% confidence interval for the mean weight of the bags is between 3.22 and 3.78 ounces. These findings suggest that consumers may be receiving less product than expected, warranting further review by the snack food company.
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