Question

Does posting calorie content for menu items affect people's choices in fast food restaurants? Accord- ing to results obtained by Elbel, Gyamfi, and Kersh (2011), the answer is no. The researchers monitored the calorie content of food purchases for children and adolescents in four large fast food chains before and after mandatory labeling began in New York City. Although most of the adolescents reported noticing the calorie labels, apparently the labels had no effect on their choices. Data similar to the results obtained show an average of M = 786 calories per meal with s = 85 for n = 100 children and adolescents before the labeling, compared to an average of M = 772 calories with s = 91 for a similar sample of n = 100 after the mandatory posting. a. Use a two-tailed test with ? = .05 to determine whether the mean number of calories after the posting is significantly different than before calorie content was posted. b. Calculate r2 to measure effect size for the mean difference.

Answer 1

Answer:

a

The decision rule is  

   Fail to reject the null hypothesis

The conclusion is  

  There is no sufficient evidence to  state  that the mean number of calories after the posting is significantly different than before calorie content was posted

b

The  effect size for the mean difference is  $r^2 = 0.0063$

Step-by-step explanation:

From the question we are told that

 The mean of calories per mean before mandatory labeling  $M_1 = 786 \ calories$

  The standard deviation is  $s_1 = 85$

   The sample size is  $n_1 = 100$

  The mean of calories per mean after mandatory labeling  $M_2 = 772$

  The standard deviation is  $s_2 = 91$

   The sample size is $n_2 = 100$

  The level of significance is $\alpha = 0.05$

The null hypothesis is  $H_o : M_1 = M_2$

The alternative hypothesis is $H_a : M_1 \ne M_2$

Generally the degree of freedom is mathematically represented as

         $df = n + n - 2$

=>      $df = 100 + 100 - 2$

=>      $df = 198$

Generally the pooled variance is mathematically represented as

         $s_p^2 = \frac{n_1 * s_1^2 + n_2 * s_2^2 }{n_1 + n_2 }$

=>     $s_p^2 = \frac{100 * 85^2 + 100 * 91^2 }{100 + 100 }$

=>     $s_p^2 = 7753$

Generally the standard error is mathematically represented as

      $SE = \sqrt{\frac{s_p^2 }{n_1} + \frac{s_p^2 }{n_2} }$

=>   $SE = \sqrt{\frac{7753 }{100} + \frac{ 7753 }{100} }$

=>   $SE = 12.45$

Generally the test statistics is mathematically represented as

        $t = \frac{ ((M_1 - M_2)-0 }{SE}$

=>     $t = \frac{ ((786 - 772 )-0 }{12.452}$

=>   $t = 1.124$

From the z t distribution table  that the probability of $( t > 1.124 )$ at a degree of freedom of  $df = 198$

    $P(t > 1.124 ) = t_{1.124 , df = 198} = 0.13118695$

Generally the p-value is mathematically represented as

      $p-value = 2 * P(t > 1.124 )$

=>   $p-value = 2 * 0.13118695$

=>   $p-value = 0.2624$

From the value obtained that we see that p-value  >  $\alpha$ so the

The decision rule is  

   Fail to reject the null hypothesis

The conclusion is  

   There is no sufficient evidence to  state  that the mean number of calories after the posting is significantly different than before calorie content was posted

Generally the value  of $r^2$ is mathematically represented as

           $r^2 = \frac{t^2 }{t^2 + df }$

=>       $r^2 = \frac{ 1.124^2 }{1.12 ^2 + 198 }$

=>       $r^2 = 0.0063$