Question

A graduate student conducted an experiment in which 22 ten-month-old babies were asked to watch a climber character attempt to scandalOn two occasions, the baby winesses the character fail to make the climb. On the third attempt the baby witness the helper toy push the character up the hill or hindrer toy prevent the character from making the scent. The helper and hinderer toys were shown to each baby in a random fashion for a fixed amount of time. The baby was then placed in front of each toy and slowed to choose which toy he or she wished to play with in 20 of the cases, the baby chose the helper toy Complete parts () through ( below (a) Why is important to randomly expose the baby to the help or hindrer toy first? OA. The randomness in the order of exposure is important to say the conditions of using the binomial probability distribution OB. The randomness in the order of exposure is important to mine the effect of the sample standard deviation OC. The rooms in the order of exposure is important to make sure half the babies so the helpen and the other half the hinderer first OD. The moderness in the order of exposure is important to avoid bias (b) What would be the appropriate ruil and automative hypotheses if the researcher is attempting to show that babies profer helpen over hinderers? Hp Vos HP Vos (c) Use the binomial probability formula to determine the value for this best P-value (Round to three decimal places as needed.) What is the correct conclusion regarding the null hypothesis? O A. Do not reject Hy. Although no level of significance is given there is insuficient evidence to suggest the proportion of babies who choose the helper toy is greater than 0.

Answer 1

The following information is provided: The sample size is N = 22, the number of favorable cases is X = 20, and the sample proportion is p= = 22 = 0.9091, and the significance level is a = 0.05 X N (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho:p=0.5 Ha: p > 0.5 This corresponds to a right-tailed test, for which a z-test for one population proportion needs to be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a right-tailed test is ze = 1.64. The rejection region for this right-tailed test is R= {z: z > 1.64}

(3) Test Statistics The 2-statistic is computed as follows: z = p – po V po(1 – po)/n 0.9091 – 0.5 10.5(1 – 0.5)/22 3.838 (4) Decision about the null hypothesis 1.64, it is then concluded that the null hypothesis is Since it is observed that z = 3.838 > 2c = rejected. - 0.0001 < 0.05, it is Using the P-value approach: The p-value is p 0.0001, and since p concluded that the null hypothesis is rejected.

(5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is greater than po, at the a = 0.05 significance level. Confidence Interval The 95% confidence interval for p is: 0.789 < p < 1.029.
C. P value = 0.0001< 0.05

And D.

d. Ans A.