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need immediate help due today!!! please check the attached pic, and please draw the bell curve. Thank you
4. A U.S. magazine, Consumer Reports, carried out a survey of the calorie and sodium content of a number of different brands of hotdogs. There were two types of hotdogs: beef and poultry. Group Beef Sample size Sample mean Sample standard deviation 20 Poultry 17 156.85 122.47 22.64 25.48 Does the poultry hotdog have a greater calorie content than the beef hotdog? What is the standard error? What is (Poult the t-test? Use three decimal places. What is the degree of freedom? What is the t-critical? (Poultry Beef) In Show Your Work, graph the bell curve. We will because . (What were we trying to prove) Show Your Work​

need immediate help due today please check the attached pic and please draw the bell curve Thank you4 A U S magazine Consumer Reports carried

Answer :

The poultry hotdog has a significantly lower calorie content. Standard error ≈ 7.990, t-test ≈ 4.300, df ≈ 43, t-critical ≈ ±2.016.

To determine whether the poultry hotdog has a greater calorie content than the beef hotdog, we can perform a two-sample t-test.

First, let's calculate the standard error (SE), the t-test statistic, the degrees of freedom (df), and the t-critical value.

1. Standard Error (SE):

[tex]\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \][/tex]

Where [tex]\( s_1 \) and \( s_2 \)[/tex] are the sample standard deviations, and [tex]\( n_1 \) and \( n_2 \)[/tex] are the sample sizes for the beef and poultry hotdogs respectively.

For the beef hotdogs:

[tex]\[ s_1 = 22.64 \]\[ n_1 = 20 \][/tex]

For the poultry hotdogs:

[tex]\[ s_2 = 25.48 \]\[ n_2 = 17 \]\[ SE = \sqrt{\frac{(22.64)^2}{20} + \frac{(25.48)^2}{17}} \]\[ SE ≈ \sqrt{\frac{511.9696}{20} + \frac{649.6304}{17}} \]\[ SE = \sqrt{25.59848 + 38.20944} \]\[ SE = \sqrt{63.80792} \]\[ SE = 7.990 \][/tex]

2. T-test statistic:

[tex]\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} \][/tex]

Where [tex]\( \bar{x}_1 \) and \( \bar{x}_2 \)[/tex] are the sample means for beef and poultry hotdogs respectively.

For the beef hotdogs:

[tex]\[ \bar{x}_1 = 156.85 \][/tex]

For the poultry hotdogs:

[tex]\[ \bar{x}_2 = 122.47 \]\[ t = \frac{156.85 - 122.47}{7.990} \]\[ t = \frac{34.38}{7.990} \]\[ t = 4.30 \][/tex]

3. Degrees of Freedom (df):

Degrees of freedom for a two-sample t-test is given by:

[tex]\[ df = \frac{(s_1^2 / n_1 + s_2^2 / n_2)^2}{\frac{(s_1^2 / n_1)^2}{n_1 - 1} + \frac{(s_2^2 / n_2)^2}{n_2 - 1}} \][/tex]

Plugging in the values:

[tex]\[ df = \frac{(22.64^2 / 20 + 25.48^2 / 17)^2}{\frac{(22.64^2 / 20)^2}{20 - 1} + \frac{(25.48^2 / 17)^2}{17 - 1}} \]\[ df = \frac{(511.9696 / 20 + 649.6304 / 17)^2}{\frac{(511.9696 / 20)^2}{19} + \frac{(649.6304 / 17)^2}{16}} \]\[ df = \frac{(25.59848 + 38.20944)^2}{\frac{648.343488}{19} + \frac{969.280768}{16}} \]\[ df = \frac{63.80792^2}{34.12176 + 60.580048} \]\[ df = \frac{4073.141571}{94.701808} \]\[ df = 43.03 \][/tex]

4. T-critical value:

For a two-tailed test with α = 0.05 and degrees of freedom ≈ 43, the t-critical value is approximately ±2.016 (you can find this value using a t-table or calculator).

Based on the calculated t-value of 4.30 and the t-critical value of ±2.016, we can conclude that the poultry hotdog indeed has a significantly greater calorie content than the beef hotdog. The calculated t-value falls in the rejection region beyond the critical value, indicating a significant difference in means.