Question

1. what causes us to overeat? one surprising factor might be the material of the plate on which our food is served. williamson, block, and keller (2016) gave n = 68 participants two donuts each and measured the amount of food that was wasted by each participant. in an independent-samples design, participants received their donuts either on a disposable paper plate or on a reusable plastic plate. data like those observed by the authors are listed below.

paper plate (grams of wasted food)	plastic plate (grams of wasted food)
35	31
34	36
36	30
40	34
34	33
33	37
39	29

a. test the hypothesis that participants who received donuts on a paper plate wasted more food than participants who were served donuts on a plastic, reusable plate. use α = .05, two - tailed.
b. construct a 95% confidence interval to estimate the size of the mean difference.
c. write the results as they would appear in a scien -

Explanation:

Step1: Calculate means and standard deviations

For Paper Plate (let's call this Group 1):
Data: 37, 35, 34, 36, 40, 34, 33, 39
$n_1 = 8$
$\bar{x}_1=\frac{37 + 35+34+36+40+34+33+39}{8}=\frac{288}{8} = 36$
$s_1=\sqrt{\frac{\sum(x_1 - \bar{x}_1)^2}{n_1 - 1}}$
Calculating deviations:
$(37 - 36)^2 = 1$, $(35 - 36)^2 = 1$, $(34 - 36)^2 = 4$, $(36 - 36)^2 = 0$, $(40 - 36)^2 = 16$, $(34 - 36)^2 = 4$, $(33 - 36)^2 = 9$, $(39 - 36)^2 = 9$
Sum of squares: $1+1 + 4+0+16+4+9+9=44$
$s_1=\sqrt{\frac{44}{7}}\approx2.507$

For Plastic Plate (Group 2):
Data: 34, 31, 36, 30, 34, 33, 37, 29
$n_2 = 8$
$\bar{x}_2=\frac{34+31+36+30+34+33+37+29}{8}=\frac{264}{8}=33$
$s_2=\sqrt{\frac{\sum(x_2 - \bar{x}_2)^2}{n_2 - 1}}$
Calculating deviations:
$(34 - 33)^2 = 1$, $(31 - 33)^2 = 4$, $(36 - 33)^2 = 9$, $(30 - 33)^2 = 9$, $(34 - 33)^2 = 1$, $(33 - 33)^2 = 0$, $(37 - 33)^2 = 16$, $(29 - 33)^2 = 16$
Sum of squares: $1+4 + 9+9+1+0+16+16=56$
$s_2=\sqrt{\frac{56}{7}}=\sqrt{8}\approx2.828$

Step2: Calculate pooled variance (for independent samples t - test)

$S_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2 - 2}=\frac{7\times(2.507)^2+7\times(2.828)^2}{14}$
$7\times6.285 + 7\times8=43.995+56 = 99.995$
$S_p^2=\frac{99.995}{14}\approx7.1425$
$S_p=\sqrt{7.1425}\approx2.672$

Step3: Calculate t - statistic

$t=\frac{\bar{x}_1-\bar{x}_2}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{36 - 33}{2.672\sqrt{\frac{1}{8}+\frac{1}{8}}}=\frac{3}{2.672\sqrt{\frac{2}{8}}}=\frac{3}{2.672\times0.5}\approx2.245$

Step4: Determine critical value and make decision (for part a)

Degrees of freedom $df=n_1 + n_2 - 2=14$
For two - tailed test with $\alpha = 0.05$, critical value $t_{crit}=\pm2.145$
Since calculated $t = 2.245>2.145$, we reject the null hypothesis.

Step5: Calculate confidence interval (for part b)

The formula for confidence interval for $\mu_1-\mu_2$ is $(\bar{x}_1-\bar{x}_2)\pm t_{\alpha/2,df}S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$
$\bar{x}_1-\bar{x}_2 = 3$
$t_{\alpha/2,14}=2.145$ (for 95% confidence, two - tailed)
$S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=2.672\times0.5 = 1.336$
Margin of error $E = 2.145\times1.336\approx2.866$
Confidence interval: $3\pm2.866=(0.134,5.866)$

Part a Answer:

We reject the null hypothesis. There is sufficient evidence at $\alpha = 0.05$ (two - tailed) to conclude that there is a difference in the amount of wasted food between paper plate and plastic plate groups. The calculated t - statistic ($t = 2.245$) is greater than the critical value ($t_{crit}=2.145$) for $df = 14$.

Part b Answer:

The 95% confidence interval for the mean difference is approximately $(0.13,5.87)$ (rounded to two decimal places).

Answer:

Step1: Calculate means and standard deviations

For Paper Plate (let's call this Group 1):
Data: 37, 35, 34, 36, 40, 34, 33, 39
$n_1 = 8$
$\bar{x}_1=\frac{37 + 35+34+36+40+34+33+39}{8}=\frac{288}{8} = 36$
$s_1=\sqrt{\frac{\sum(x_1 - \bar{x}_1)^2}{n_1 - 1}}$
Calculating deviations:
$(37 - 36)^2 = 1$, $(35 - 36)^2 = 1$, $(34 - 36)^2 = 4$, $(36 - 36)^2 = 0$, $(40 - 36)^2 = 16$, $(34 - 36)^2 = 4$, $(33 - 36)^2 = 9$, $(39 - 36)^2 = 9$
Sum of squares: $1+1 + 4+0+16+4+9+9=44$
$s_1=\sqrt{\frac{44}{7}}\approx2.507$

For Plastic Plate (Group 2):
Data: 34, 31, 36, 30, 34, 33, 37, 29
$n_2 = 8$
$\bar{x}_2=\frac{34+31+36+30+34+33+37+29}{8}=\frac{264}{8}=33$
$s_2=\sqrt{\frac{\sum(x_2 - \bar{x}_2)^2}{n_2 - 1}}$
Calculating deviations:
$(34 - 33)^2 = 1$, $(31 - 33)^2 = 4$, $(36 - 33)^2 = 9$, $(30 - 33)^2 = 9$, $(34 - 33)^2 = 1$, $(33 - 33)^2 = 0$, $(37 - 33)^2 = 16$, $(29 - 33)^2 = 16$
Sum of squares: $1+4 + 9+9+1+0+16+16=56$
$s_2=\sqrt{\frac{56}{7}}=\sqrt{8}\approx2.828$

Step2: Calculate pooled variance (for independent samples t - test)

$S_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2 - 2}=\frac{7\times(2.507)^2+7\times(2.828)^2}{14}$
$7\times6.285 + 7\times8=43.995+56 = 99.995$
$S_p^2=\frac{99.995}{14}\approx7.1425$
$S_p=\sqrt{7.1425}\approx2.672$

Step3: Calculate t - statistic

$t=\frac{\bar{x}_1-\bar{x}_2}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{36 - 33}{2.672\sqrt{\frac{1}{8}+\frac{1}{8}}}=\frac{3}{2.672\sqrt{\frac{2}{8}}}=\frac{3}{2.672\times0.5}\approx2.245$

Step4: Determine critical value and make decision (for part a)

Degrees of freedom $df=n_1 + n_2 - 2=14$
For two - tailed test with $\alpha = 0.05$, critical value $t_{crit}=\pm2.145$
Since calculated $t = 2.245>2.145$, we reject the null hypothesis.

Step5: Calculate confidence interval (for part b)

The formula for confidence interval for $\mu_1-\mu_2$ is $(\bar{x}_1-\bar{x}_2)\pm t_{\alpha/2,df}S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$
$\bar{x}_1-\bar{x}_2 = 3$
$t_{\alpha/2,14}=2.145$ (for 95% confidence, two - tailed)
$S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=2.672\times0.5 = 1.336$
Margin of error $E = 2.145\times1.336\approx2.866$
Confidence interval: $3\pm2.866=(0.134,5.866)$

Part a Answer:

We reject the null hypothesis. There is sufficient evidence at $\alpha = 0.05$ (two - tailed) to conclude that there is a difference in the amount of wasted food between paper plate and plastic plate groups. The calculated t - statistic ($t = 2.245$) is greater than the critical value ($t_{crit}=2.145$) for $df = 14$.

Part b Answer:

The 95% confidence interval for the mean difference is approximately $(0.13,5.87)$ (rounded to two decimal places).