Math  /  Data & Statistics

Question4. The online security firm SecurEnvoy and the research firm OnePoll conducted a survey in October 2012 and found that 60%60 \% of Britons admit they don't always understand text message abbreviations they receive. The firms surveyed 1000 British adults. a) Assuming this was a simple random sample, can you be comfortable that 60%60 \% is a good estimate of the percentage of British adults who are sometimes confused by text abbreviations? Explain. b) Assuming this was a simple random sample, can you be comfortable that 60%60 \% is a good estimate of the percentage of American adults who are sometimes confused by text abbreviations? Explain. c) It seems reasonable to suspect that age may be associated with a person's comfort with text abbreviations. How might the sampling technique be improved by taking this association into account? d) A blog that reported a story about this poll had a banner at the bottom of the webpage with a multiple choice question: Do you get confused with abbreviations in text messages? - Yes. Sometimes the abbreviations do the opposite of what they're supposed to do. - No. I've been texting for a long time. It's my second language - IDK (tsminteractive.com/have-you-ever-been-confused-by-text-mes-sage-abbreviations-poll/)
What type of sampling is the blog using? Will the results of their survey be likely to match those of the original survey? Explain.

Studdy Solution

STEP 1

1. The survey conducted by SecurEnvoy and OnePoll is assumed to be a simple random sample of 1000 British adults.
2. The percentage reported is 60%60\%.
3. The blog uses a different sampling method, which needs to be identified and analyzed.

STEP 2

1. Evaluate the reliability of the 60%60\% estimate for British adults.
2. Evaluate the reliability of the 60%60\% estimate for American adults.
3. Suggest improvements in sampling technique considering age association.
4. Identify the sampling method used by the blog and compare it with the original survey.

STEP 3

a) To determine if 60%60\% is a good estimate for British adults, consider the sample size and sampling method. A simple random sample of 1000 is typically sufficient for estimating proportions in a large population, assuming the sample is representative. The margin of error can be calculated using the formula for the confidence interval of a proportion:
Margin of Error=zp(1p)n\text{Margin of Error} = z \cdot \sqrt{\frac{p(1-p)}{n}}
where z z is the z-score for the desired confidence level (commonly 1.96 for 95%95\%), p=0.60 p = 0.60 , and n=1000 n = 1000 .
Calculate the margin of error:
Margin of Error=1.960.60×0.4010000.03\text{Margin of Error} = 1.96 \cdot \sqrt{\frac{0.60 \times 0.40}{1000}} \approx 0.03
This means the true proportion is likely between 57%57\% and 63%63\%. Given this margin, 60%60\% is a reasonable estimate for British adults.

STEP 4

b) To determine if 60%60\% is a good estimate for American adults, consider cultural and demographic differences. The sample was taken from British adults, and there may be significant differences in texting habits and language use between British and American populations. Therefore, without additional data, 60%60\% is not a reliable estimate for American adults.

STEP 5

c) To improve the sampling technique considering age, stratified sampling could be used. This involves dividing the population into age groups (strata) and sampling from each group proportionally. This ensures that all age groups are represented in the sample, which may provide a more accurate estimate of the overall population's comfort with text abbreviations.

STEP 6

d) The blog is likely using a voluntary response sampling method, where individuals choose to participate. This method often leads to bias, as those who feel strongly about the topic are more likely to respond. The results are unlikely to match the original survey, which used a simple random sample, because the blog's sample is not representative of the general population.

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