Question

several magazines were grouped into three categories based on what level of education of their readers the magazines are geared towards: high, medium, or low level. then random samples of the magazines were selected to determine the number of three - plus - syllable words were in the advertising copy, and the data is in the table. is there enough evidence to show that the mean number of three - plus - syllable words in advertising copy is different for at least one of the education levels? test at the 5% level.
number of three plus syllable words in advertising copy
state the hypotheses.
( h_0 ): select an answer
( h_a ): select an answer
calculate the ( f ) test statistic. round to two decimal places.
( f=)
calculate the p - value. round to four decimal places.
( p - value=)
state your decision.
since p - value ( quad 0.05 ), select an answer ( h_0 ).
interpret the results.
at the 5% significance level, there select an answer evidence to show that

Explanation:

Step1: State the hypotheses

The null hypothesis \(H_0\) is that the mean number of three - plus - syllable words is the same for all education levels (\(\mu_{High}=\mu_{Medium}=\mu_{Low}\)). The alternative hypothesis \(H_a\) is that at least one mean is different.

Step2: Calculate the group means and overall mean

Let \(n_1 = n_2=n_3 = 20\) (since each group has 20 data points).
For the high - education group:
\(\bar{x}_1=\frac{42 + 39+22+\cdots+28}{20}=\frac{633}{20}=31.65\)
For the medium - education group:
\(\bar{x}_2=\frac{39+34 + 31+\cdots+30}{20}=\frac{540}{20}=27\)
For the low - education group:
\(\bar{x}_3=\frac{9+11+17+\cdots+9}{20}=\frac{270}{20}=13.5\)
The overall mean \(\bar{x}=\frac{633 + 540+270}{20\times3}=\frac{1443}{60}=24.05\)

Step3: Calculate the sum of squares between groups (\(SSB\))

\(SSB=n_1(\bar{x}_1-\bar{x})^2 + n_2(\bar{x}_2-\bar{x})^2+n_3(\bar{x}_3-\bar{x})^2\)
\(SSB = 20\times(31.65 - 24.05)^2+20\times(27 - 24.05)^2+20\times(13.5 - 24.05)^2\)
\(SSB=20\times(7.6)^2+20\times(2.95)^2+20\times(- 10.55)^2\)
\(SSB=20\times(57.76+8.7025 + 111.3025)\)
\(SSB=20\times177.765=3555.3\)

Step4: Calculate the sum of squares within groups (\(SSW\))

For the high - education group:
\(SSW_1=\sum_{i = 1}^{20}(x_{i1}-\bar{x}_1)^2=(42 - 31.65)^2+(39 - 31.65)^2+\cdots+(28 - 31.65)^2=1377.15\)
For the medium - education group:
\(SSW_2=\sum_{i = 1}^{20}(x_{i2}-\bar{x}_2)^2=(39 - 27)^2+(34 - 27)^2+\cdots+(30 - 27)^2=1074\)
For the low - education group:
\(SSW_3=\sum_{i = 1}^{20}(x_{i3}-\bar{x}_3)^2=(9 - 13.5)^2+(11 - 13.5)^2+\cdots+(9 - 13.5)^2=1791\)
\(SSW=SSW_1+SSW_2+SSW_3=1377.15 + 1074+1791=4242.15\)

Step5: Calculate the degrees of freedom

Degrees of freedom between groups \(df_1=k - 1=3 - 1 = 2\) (where \(k = 3\) groups)
Degrees of freedom within groups \(df_2=n - k=60 - 3=57\)

Step6: Calculate the mean square between groups (\(MSB\)) and mean square within groups (\(MSW\))

\(MSB=\frac{SSB}{df_1}=\frac{3555.3}{2}=1777.65\)
\(MSW=\frac{SSW}{df_2}=\frac{4242.15}{57}\approx74.42\)

Step7: Calculate the \(F\) test statistic

\(F=\frac{MSB}{MSW}=\frac{1777.65}{74.42}\approx23.89\)

Step8: Calculate the \(p\) - value

Using an \(F\) - distribution table or software with \(df_1 = 2\) and \(df_2=57\), the \(p\) - value \(P(F>23.89)\approx0.0000\) (using statistical software like R: 1 - pf(23.89,2,57) gives a very small value close to \(0\))

Step9: Make a decision

Since \(p - value\approx0.0000<0.05\), we reject \(H_0\)

Answer:

\(H_0:\mu_{High}=\mu_{Medium}=\mu_{Low}\), \(H_a:\) At least one \(\mu_i\) is different.
\(F = 23.89\)
\(p - value\approx0.0000\)
Since \(p - value<0.05\), reject \(H_0\)
At the \(5\%\) significance level, there is sufficient evidence to show that the mean number of three - plus - syllable words in advertising copy is different for at least one of the education levels.