A survey for brand recognition is done and it is determined that 68% of consumers have heard of Dull Computer
Company. A survey of 800 randomly selected consumers is to be conducted. For such groups of 800, would it be unusual
to get 634 consumers who recognize the Dull Computer Company name?
To determine if it would be unusual to get 634 consumers who recognize the Dull Computer Company name out of a group of 800, we can use the concept of standard deviation and the normal distribution.
First, calculate the standard deviation (σ) using the following formula:
σ = √(p * (1 - p) / n),
where p is the proportion of consumers who recognize the Dull Computer Company name (0.68 in this case) and n is the sample size (800).
σ = √(0.68 * (1 - 0.68) / 800)
= √(0.2176 / 800)
= √0.000272
≈ 0.0165
Next, we can calculate the z-score using the formula:
z = (x - μ) / σ,
where x is the observed number of consumers who recognize the brand (634 in this case), μ is the expected number of consumers who recognize the brand (0.68 * 800 = 544), and σ is the standard deviation calculated above.
z = (634 - 544) / 0.0165
= 5424 / 0.0165
≈ 328727.27
The z-score represents the number of standard deviations away from the expected value. In this case, the z-score is quite large (328727.27), indicating that the observed number of consumers who recognize the brand is significantly different from the expected number.
Now, we can determine if the result is unusual by looking up the z-score in a standard normal distribution table (or by using a statistical calculator).
For a standard normal distribution, a z-score that falls beyond ±1.96 standard deviations from the mean is considered unusual. In this case, the z-score of 328727.27 is way beyond this threshold, indicating that getting 634 consumers who recognize the Dull Computer Company name out of 800 would be highly unusual.
Therefore, it would be unusual to get 634 consumers who recognize the Dull Computer Company name out of a group of 800 based on the given proportion of brand recognition.