(4a^2)b–ab^2–(3a^2)b+ab^2–ab+6 for a=−3, b=2
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Answer:
We conclude that a negative message results in a lower mean score than positive message.
Step-by-step 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.
<em>Let </em><em> = population mean score for negative message</em>
<em /><em> = population mean score for positive message</em>
SO, Null Hypothesis, :
or
{means that a negative message results in a higher or equal mean score than positive message}
Alternate Hypothesis, :
or
{means that a negative message results in a lower mean score than positive message}
The test statistics that will be used here is <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;
T.S. = ~
where, = sample mean score for negative message = 9.64
= sample mean score for positive message = 15.84
= sample standard deviation for negative message = 3.43
= sample standard deviation for positive message = 8.65
= sample of subjects receiving the negative message = 21
= sample of subjects receiving the positive message = 21
Also, =
= 6.58
So, <u><em>the test statistics</em></u> = ~
= -3.053
<em>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.</em>
Therefore, we conclude that a negative message results in a lower mean score than positive message.