Question

Do positive or negative messages have a greater effect on behavior? Forty-two subjects were randomly assigned to one of two treatment groups, 21 per group. In one group, subjects read an introduction about a nonprofit organization followed by a negative message encouraging subjects to sign a petition for cleaner lakes. In the other group, subjects read the same introduction followed by a positive message encouraging subjects to sign the petition. All subjects were then tested to determine how strongly they intended to sign the petition, with higher scores indicating greater intent. The 21 subjects receiving the negative message had a mean score of 9.64 with standard deviation 3.43; the 21 subjects receiving the positive message had a mean score of 15.84 with standard deviation 8.65.

At ???? = 0.05, is there enough evidence to say that a negative message results in a lower mean score?

Answer 1

We conclude that a negative message results in a lower mean score than positive message.

Explanation:

We are given that Forty-two subjects were randomly assigned to one of two treatment groups, 21 per group.

The 21 subjects receiving the negative message had a mean score of 9.64 with standard deviation 3.43; the 21 subjects receiving the positive message had a mean score of 15.84 with standard deviation 8.65.

Let
`\mu_1` = population mean score for negative message

`\mu_2` = population mean score for positive message

SO, Null Hypothesis,
`H_0` :
`\mu_1-\mu_2\geq0` or
`\mu_1\geq \mu_2` {means that a negative message results in a higher or equal mean score than positive message}

Alternate Hypothesis,
`H_A` :
`\mu_1-\mu_2<0` or
`\mu_1<\mu_2` {means that a negative message results in a lower mean score than positive message}

The test statistics that will be used here is Two-sample t test statistics as we don't know about the population standard deviations;

T.S. =
`\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{(1)/(n_1)+(1)/(n_2) } }` ~
`t__n__1+_n__2-2`

where,
`\bar X_1` = sample mean score for negative message = 9.64

`\bar X_2` = sample mean score for positive message = 15.84

`s_1` = sample standard deviation for negative message = 3.43

`s_2` = sample standard deviation for positive message = 8.65

`n_1` = sample of subjects receiving the negative message = 21

`n_2` = sample of subjects receiving the positive message = 21

Also,
`s_p=\sqrt{((n_1-1)s_1^(2)+(n_2-1)s_2^(2) )/(n_1+n_2-2) }` =
`\sqrt{((21-1)* 3.43^(2)+(21-1)* 8.65^(2) )/(21+21-2) }` = 6.58

So, the test statistics =
`\frac{(9.64-15.84)-(0)}{6.58 * \sqrt{(1)/(21)+(1)/(21) } }` ~
`t_4_0`

= -3.053

Now at 0.05 significance level, the t table gives critical value of -1.684 at 40 degree of freedom for left-tailed test. Since our test statistics is less than the critical value of t as -3.053 < -1.684, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that a negative message results in a lower mean score than positive message.